I'm trying to transfer my understanding of plyr into dplyr, but I can't figure out how to group by multiple columns.
# make data with weird column names that can't be hard coded
data = data.frame(
asihckhdoydkhxiydfgfTgdsx = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkghc5cdsvxvyv0ja = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
# get the columns we want to average within
columns = names(data)[-3]
# plyr - works
ddply(data, columns, summarize, value=mean(value))
# dplyr - raises error
data %.%
group_by(columns) %.%
summarise(Value = mean(value))
#> Error in eval(expr, envir, enclos) : index out of bounds
What am I missing to translate the plyr example into a dplyr-esque syntax?
Edit 2017: Dplyr has been updated, so a simpler solution is available. See the currently selected answer.
Just so as to write the code in full, here's an update on Hadley's answer with the new syntax:
library(dplyr)
df <- data.frame(
asihckhdoydk = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkgh = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
# Columns you want to group by
grp_cols <- names(df)[-3]
# Convert character vector to list of symbols
dots <- lapply(grp_cols, as.symbol)
# Perform frequency counts
df %>%
group_by_(.dots=dots) %>%
summarise(n = n())
output:
Source: local data frame [9 x 3]
Groups: asihckhdoydk
asihckhdoydk a30mvxigxkgh n
1 A A 10
2 A B 10
3 A C 13
4 B A 14
5 B B 10
6 B C 12
7 C A 9
8 C B 12
9 C C 10
Since this question was posted, dplyr added scoped versions of group_by (documentation here). This lets you use the same functions you would use with select, like so:
data = data.frame(
asihckhdoydkhxiydfgfTgdsx = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkghc5cdsvxvyv0ja = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
# get the columns we want to average within
columns = names(data)[-3]
library(dplyr)
df1 <- data %>%
group_by_at(vars(one_of(columns))) %>%
summarize(Value = mean(value))
#compare plyr for reference
df2 <- plyr::ddply(data, columns, plyr::summarize, value=mean(value))
table(df1 == df2, useNA = 'ifany')
## TRUE
## 27
The output from your example question is as expected (see comparison to plyr above and output below):
# A tibble: 9 x 3
# Groups: asihckhdoydkhxiydfgfTgdsx [?]
asihckhdoydkhxiydfgfTgdsx a30mvxigxkghc5cdsvxvyv0ja Value
<fctr> <fctr> <dbl>
1 A A 0.04095002
2 A B 0.24943935
3 A C -0.25783892
4 B A 0.15161805
5 B B 0.27189974
6 B C 0.20858897
7 C A 0.19502221
8 C B 0.56837548
9 C C -0.22682998
Note that since dplyr::summarize only strips off one layer of grouping at a time, you've still got some grouping going on in the resultant tibble (which can sometime catch people by suprise later down the line). If you want to be absolutely safe from unexpected grouping behavior, you can always add %>% ungroup to your pipeline after you summarize.
The support for this in dplyr is currently pretty weak, eventually I think the syntax will be something like:
df %.% group_by(.groups = c("asdfgfTgdsx", "asdfk30v0ja"))
But that probably won't be there for a while (because I need to think through all the consequences).
In the meantime, you can use regroup(), which takes a list of symbols:
library(dplyr)
df <- data.frame(
asihckhdoydk = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkgh = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
df %.%
regroup(list(quote(asihckhdoydk), quote(a30mvxigxkgh))) %.%
summarise(n = n())
If you have have a character vector of column names, you can convert them to the right structure with lapply() and as.symbol():
vars <- setdiff(names(df), "value")
vars2 <- lapply(vars, as.symbol)
df %.% regroup(vars2) %.% summarise(n = n())
String specification of columns in dplyr are now supported through variants of the dplyr functions with names finishing in an underscore. For example, corresponding to the group_by function there is a group_by_ function that may take string arguments. This vignette describes the syntax of these functions in detail.
The following snippet cleanly solves the problem that #sharoz originally posed (note the need to write out the .dots argument):
# Given data and columns from the OP
data %>%
group_by_(.dots = columns) %>%
summarise(Value = mean(value))
(Note that dplyr now uses the %>% operator, and %.% is deprecated).
Update with across() from dplyr 1.0.0
All the answers above are still working, and the solutions with the .dots argument are intruiging.
BUT if you look for a solution that is easier to remember, the new across() comes in handy. It was published 2020-04-03 by Hadley Wickham and can be used in mutate() and summarise() and replace the scoped variants like _at or _all. Above all, it replaces very elegantly the cumbersome non-standard evaluation (NSE) with quoting/unquoting such as !!! rlang::syms().
So the solution with across looks very readable:
data %>%
group_by(across(all_of(columns))) %>%
summarize(Value = mean(value))
Until dplyr has full support for string arguments, perhaps this gist is useful:
https://gist.github.com/skranz/9681509
It contains bunch of wrapper functions like s_group_by, s_mutate, s_filter, etc that use string arguments. You can mix them with the normal dplyr functions. For example
cols = c("cyl","gear")
mtcars %.%
s_group_by(cols) %.%
s_summarise("avdisp=mean(disp), max(disp)") %.%
arrange(avdisp)
It works if you pass it the objects (well, you aren't, but...) rather than as a character vector:
df %.%
group_by(asdfgfTgdsx, asdfk30v0ja) %.%
summarise(Value = mean(value))
> df %.%
+ group_by(asdfgfTgdsx, asdfk30v0ja) %.%
+ summarise(Value = mean(value))
Source: local data frame [9 x 3]
Groups: asdfgfTgdsx
asdfgfTgdsx asdfk30v0ja Value
1 A C 0.046538002
2 C B -0.286359899
3 B A -0.305159419
4 C A -0.004741504
5 B B 0.520126476
6 C C 0.086805492
7 B C -0.052613078
8 A A 0.368410146
9 A B 0.088462212
where df was your data.
?group_by says:
...: variables to group by. All tbls accept variable names, some
will also accept functons of variables. Duplicated groups
will be silently dropped.
which I interpret to mean not the character versions of the names, but how you would refer to them in foo$bar; bar is not quoted here. Or how you'd refer to variables in a formula: foo ~ bar.
#Arun also mentions that you can do:
df %.%
group_by("asdfgfTgdsx", "asdfk30v0ja") %.%
summarise(Value = mean(value))
But you can't pass in something that unevaluated is not a name of a variable in the data object.
I presume this is due to the internal methods Hadley is using to look up the things you pass in via the ... argument.
data = data.frame(
my.a = sample(LETTERS[1:3], 100, replace=TRUE),
my.b = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
group_by(data,newcol=paste(my.a,my.b,sep="_")) %>% summarise(Value=mean(value))
One (tiny) case that is missing from the answers here, that I wanted to make explicit, is when the variables to group by are generated dynamically midstream in a pipeline:
library(wakefield)
df_foo = r_series(rnorm, 10, 1000)
df_foo %>%
# 1. create quantized versions of base variables
mutate_each(
funs(Quantized = . > 0)
) %>%
# 2. group_by the indicator variables
group_by_(
.dots = grep("Quantized", names(.), value = TRUE)
) %>%
# 3. summarize the base variables
summarize_each(
funs(sum(., na.rm = TRUE)), contains("X_")
)
This basically shows how to use grep in conjunction with group_by_(.dots = ...) to achieve this.
General example on using the .dots argument as character vector input to the dplyr::group_by function :
iris %>%
group_by(.dots ="Species") %>%
summarise(meanpetallength = mean(Petal.Length))
Or without a hard coded name for the grouping variable (as asked by the OP):
iris %>%
group_by(.dots = names(iris)[5]) %>%
summarise_at("Petal.Length", mean)
With the example of the OP:
data %>%
group_by(.dots =names(data)[-3]) %>%
summarise_at("value", mean)
See also the dplyr vignette on programming which explains pronouns, quasiquotation, quosures, and tidyeval.
Related
I would like to create for loop to repeat the same function for 150 variables. I am new to R and I am a bit stuck.
To give you an example of some commands I need to repeat:
N <- table(df$ var1 ==0)["TRUE"]
n <- table(df$ var1 ==1)["TRUE"]
PREV95 <- (svyciprop(~ var1 ==1, level=0.95, design= design, deff= "replace")*100)
I need to run the same functions for 150 columns. I know that I need to put all my cols in one vector = x but then I don't know how to write the loop to repeat the same command for all my variables.
Can anyone help me to write a loop?
A word in advance: loops in R can in most cases be replaced with a faster, R-ish way (various flavours of apply, maping, walking ...)
applying a function to the columns of dataframe df:
a)
with base R, example dataset cars
my_function <- function(xs) max(xs)
lapply(cars, my_function)
b)
tidyverse-style:
cars %>%
summarise_all(my_function)
An anecdotal example: I came across an R-script which took about half an hour to complete and made abundant use of for-loops. Replacing the loops with vectorized functions and members of the apply family cut the execution time down to about 3 minutes. So while for-loops and related constructs might be more familiar when coming from another language, they might soon get in your way with R.
This chapter of Hadley Wickham's R for data science gives an introduction into iterating "the R-way".
Here is an approach that doesn't use loops. I've created a data set called df with three factor variables to represent your dataset as you described it. I created a function eval() that does all the work. First, it filters out just the factors. Then it converts your factors to numeric variables so that the numbers can be summed as 0 and 1 otherwise if we sum the factors it would be based on 1 and 2. Within the function I create another function neg() to give you the number of negative values by subtracting the sum of the 1s from the total length of the vector. Then create the dataframes "n" (sum of the positives), "N" (sum of the negatives), and PREV95. I used pivot_longer to get the data in a long format so that each stat you are looking for will be in its own column when merged together. Note I had to leave PREV95 out because I do not have a 'design' object to use as a parameter to run the function. I hashed it out but you can remove the hash to add back in. I then used left_join to combine these dataframes and return "results". Again, I've hashed out the version that you'd use to include PREV95. The function eval() takes your original dataframe as input. I think the logic for PREV95 should work, but I cannot check it without a 'design' parameter. It returns a dataframe, not a list, which you'll likely find easier to work with.
library(dplyr)
library(tidyr)
seed(100)
df <- data.frame(Var1 = factor(sample(c(0,1), 10, TRUE)),
Var2 = factor(sample(c(0,1), 10, TRUE)),
Var3 = factor(sample(c(0,1), 10, TRUE)))
eval <- function(df){
df1 <- df %>%
select_if(is.factor) %>%
mutate_all(function(x) as.numeric(as.character(x)))
neg <- function(x){
length(x) - sum(x)
}
n<- df1 %>%
summarize(across(where(is.numeric), sum)) %>%
pivot_longer(everything(), names_to = "Var", values_to = "n")
N <- df1 %>%
summarize(across(where(is.numeric), function(x) neg(x))) %>%
pivot_longer(everything(), names_to = "Var", values_to = "N")
#PREV95 <- df1 %>%
# summarize(across(where(is.numeric), function(x) survey::svyciprop(~x == 1, design = design, level = 0.95, deff = "replace")*100)) %>%
# pivot_longer(everything(), names_to = "Var", values_to = "PREV95")
results <- n %>%
left_join(N, by = "Var")
#results <- n %>%
# left_join(N, by = "Var") %>%
# left_join(PREV95, by = "Var")
return(results)
}
eval(df)
Var n N
<chr> <dbl> <dbl>
1 Var1 2 8
2 Var2 5 5
3 Var3 4 6
If you really wanted to use a for loop, here is how to make it work. Again, I've left out the survey function due to a lack of info on the parameters to make it work.
seed(100)
df <- data.frame(Var1 = factor(sample(c(0,1), 10, TRUE)),
Var2 = factor(sample(c(0,1), 10, TRUE)),
Var3 = factor(sample(c(0,1), 10, TRUE)))
VarList <- names(df %>% select_if(is.factor))
results <- list()
for (var in VarList){
results[[var]][["n"]] <- sum(df[[var]] == 1)
results[[var]][["N"]] <- sum(df[[var]] == 0)
}
unlist(results)
Var1.n Var1.N Var2.n Var2.N Var3.n Var3.N
2 8 5 5 4 6
Well, I know that there are already tons of related questions, but none gave an answer to my particular need.
I want to use dplyr "summarize" on a table with 50 columns, and I need to apply different summary functions to these.
"Summarize_all" and "summarize_at" both seem to have the disadvantage that it's not possible to apply different functions to different subgroups of variables.
As an example, let's assume the iris dataset would have 50 columns, so we do not want to address columns by names. I want the sum over the first two columns, the mean over the third and the first value for all remaining columns (after a group_by(Species)). How could I do this?
Fortunately, there is a much simpler way available now.
With the new dplyr 1.0.0 coming out soon, you can leverage the across function for this purpose.
All you need to type is:
iris %>%
group_by(Species) %>%
summarize(
# I want the sum over the first two columns,
across(c(1,2), sum),
# the mean over the third
across(3, mean),
# the first value for all remaining columns (after a group_by(Species))
across(-c(1:3), first)
)
Great, isn't it?
I first thought the across is not necessary as the scoped variants worked just fine, but this use case is exactly why the across function can be very beneficial.
You can get the latest version of dplyr by devtools::install_github("tidyverse/dplyr")
As other people have mentioned, this is normally done by calling summarize_each / summarize_at / summarize_if for every group of columns that you want to apply the summarizing function to. As far as I know, you would have to create a custom function that performs summarizations to each subset. You can for example set the colnames in such way that you can use the select helpers (e.g. contains()) to filter just the columns that you want to apply the function to. If not, then you can set the specific column numbers that you want to summarize.
For the example you mentioned, you could try the following:
summarizer <- function(tb, colsone, colstwo, colsthree,
funsone, funstwo, funsthree, group_name) {
return(bind_cols(
summarize_all(select(tb, colsone), .funs = funsone),
summarize_all(select(tb, colstwo), .funs = funstwo) %>%
ungroup() %>% select(-matches(group_name)),
summarize_all(select(tb, colsthree), .funs = funsthree) %>%
ungroup() %>% select(-matches(group_name))
))
}
#With colnames
iris %>% as.tibble() %>%
group_by(Species) %>%
summarizer(colsone = contains("Sepal"),
colstwo = matches("Petal.Length"),
colsthree = c(-contains("Sepal"), -matches("Petal.Length")),
funsone = "sum",
funstwo = "mean",
funsthree = "first",
group_name = "Species")
#With indexes
iris %>% as.tibble() %>%
group_by(Species) %>%
summarizer(colsone = 1:2,
colstwo = 3,
colsthree = 4,
funsone = "sum",
funstwo = "mean",
funsthree = "first",
group_name = "Species")
You could summarise the data with each function separately and then join the data later if needed.
So something like this for the iris example:
sums <- iris %>% group_by(Species) %>% summarise_at(1:2, sum)
means <- iris %>% group_by(Species) %>% summarise_at(3, mean)
firsts <- iris %>% group_by(Species) %>% summarise_at(4, first)
full_join(sums, means) %>% full_join(firsts)
Though I would try to think of something else if there are more than a handful of summarising functions you need to use.
Try this:
library(plyr)
library(dplyr)
dataframe <- data.frame(var = c(1,1,1,2,2,2),var2 = c(10,9,8,7,6,5),var3=c(2,3,4,5,6,7),var4=c(5,5,3,2,4,2))
dataframe
# var var2 var3 var4
#1 1 10 2 5
#2 1 9 3 5
#3 1 8 4 3
#4 2 7 5 2
#5 2 6 6 4
#6 2 5 7 2
funnames<-c(sum,mean,first)
colnums<-c(2,3,4)
ddply(.data = dataframe,.variables = "var",
function(x,funcs,inds){
mapply(function(func,ind){
func(x[,ind])
},funcs,inds)
},funnames,colnums)
# var V1 V2 V3
#1 1 27 3 5
#2 2 18 6 2
See this - feature coming soon
I am trying to use pipe mutate statement using a custom function. I looked a this somewhat similar SO post but in vain.
Say I have a data frame like this (where blob is some variable not related to the specific task but is part of the entire data) :
df <-
data.frame(exclude=c('B','B','D'),
B=c(1,0,0),
C=c(3,4,9),
D=c(1,1,0),
blob=c('fd', 'fs', 'sa'),
stringsAsFactors = F)
I have a function that uses the variable names so select some based on the value in the exclude column and e.g. calculates a sum on the variables not specified in exclude (which is always a single character).
FUN <- function(df){
sum(df[c('B', 'C', 'D')] [!names(df[c('B', 'C', 'D')]) %in% df['exclude']] )
}
When I gives a single row (row 1) to FUN I get the the expected sum of C and D (those not mentioned by exclude), namely 4:
FUN(df[1,])
How do I do similarly in a pipe with mutate (adding the result to a variable s). These two tries do not work:
df %>% mutate(s=FUN(.))
df %>% group_by(1:n()) %>% mutate(s=FUN(.))
UPDATE
This also do not work as intended:
df %>% rowwise(.) %>% mutate(s=FUN(.))
This works of cause but is not within dplyr's mutate (and pipes):
df$s <- sapply(1:nrow(df), function(x) FUN(df[x,]))
If you want to use dplyr you can do so using rowwise and your function FUN.
df %>%
rowwise %>%
do({
result = as_data_frame(.)
result$s = FUN(result)
result
})
The same can be achieved using group_by instead of rowwise (like you already tried) but with do instead of mutate
df %>%
group_by(1:n()) %>%
do({
result = as_data_frame(.)
result$s = FUN(result)
result
})
The reason mutate doesn't work in this case, is that you are passing the whole tibble to it, so it's like calling FUN(df).
A much more efficient way of doing the same thing though is to just make a matrix of columns to be included and then use rowSums.
cols <- c('B', 'C', 'D')
include_mat <- outer(function(x, y) x != y, X = df$exclude, Y = cols)
# or outer(`!=`, X = df$exclude, Y = cols) if it's more readable to you
df$s <- rowSums(df[cols] * include_mat)
purrr approach
We can use a combination of nest and map_dbl for this:
library(tidyverse)
df %>%
rowwise %>%
nest(-blob) %>%
mutate(s = map_dbl(data, FUN)) %>%
unnest
Let's break that down a little bit. First, rowwise allows us to apply each subsequent function to support arbitrary complex operations that need to be applied to each row.
Next, nest will create a new column that is a list of our data to be fed into FUN (the beauty of tibbles vs data.frames!). Since we are applying this rowwise, each row contains a single-row tibble of exclude:D.
Finally, we use map_dbl to map our FUN to each of these tibbles. map_dbl is used over the family of other map_* functions since our intended output is numeric (i.e. double).
unnest returns our tibble into the more standard structure.
purrrlyr approach
While purrrlyr may not be as 'popular' as its parents dplyr and purrr, its by_row function has some utility here.
In your above example, we would use your data frame df and user-defined function FUN in the following way:
df %>%
by_row(..f = FUN, .to = "s", .collate = "cols")
That's it! Giving you:
# tibble [3 x 6]
exclude B C D blob s
<chr> <dbl> <dbl> <dbl> <chr> <dbl>
1 B 1 3 1 fd 4
2 B 0 4 1 fs 5
3 D 0 9 0 sa 9
Admittedly, the syntax is a little strange, but here's how it breaks down:
..f = the function to apply to each row
.to = the name of the output column, in this case s
.collate = the way the results should be collated, by list, row, or column. Since FUN only has a single output, we would be fine to use either "cols" or "rows"
See here for more information on using purrrlyr...
Performance
Forewarning, while I like the functionality of by_row, it's not always the best approach for performance! purrr is more intuitive, but also at a rather large speed loss. See the following microbenchmark test:
library(microbenchmark)
mbm <- microbenchmark(
purrr.test = df %>% rowwise %>% nest(-blob) %>%
mutate(s = map_dbl(data, FUN)) %>% unnest,
purrrlyr.test = df %>% by_row(..f = FUN, .to = "s", .collate = "cols"),
rowwise.test = df %>%
rowwise %>%
do({
result = as_tibble(.)
result$s = FUN(result)
result
}),
group_by.test = df %>%
group_by(1:n()) %>%
do({
result = as_tibble(.)
result$s = FUN(result)
result
}),
sapply.test = {df$s <- sapply(1:nrow(df), function(x) FUN(df[x,]))},
times = 1000
)
autoplot(mbm)
You can see that the purrrlyr approach is faster than the approach of using a combination of do with rowwise or group_by(1:n()) (see #konvas answer), and rather on par with the sapply approach. However, the package is admittedly not the most intuitive. The standard purrr approach seems to be the slowest, but also perhaps easier to work with. Different user-defined functions may change the speed order.
I'm trying to transfer my understanding of plyr into dplyr, but I can't figure out how to group by multiple columns.
# make data with weird column names that can't be hard coded
data = data.frame(
asihckhdoydkhxiydfgfTgdsx = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkghc5cdsvxvyv0ja = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
# get the columns we want to average within
columns = names(data)[-3]
# plyr - works
ddply(data, columns, summarize, value=mean(value))
# dplyr - raises error
data %.%
group_by(columns) %.%
summarise(Value = mean(value))
#> Error in eval(expr, envir, enclos) : index out of bounds
What am I missing to translate the plyr example into a dplyr-esque syntax?
Edit 2017: Dplyr has been updated, so a simpler solution is available. See the currently selected answer.
Just so as to write the code in full, here's an update on Hadley's answer with the new syntax:
library(dplyr)
df <- data.frame(
asihckhdoydk = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkgh = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
# Columns you want to group by
grp_cols <- names(df)[-3]
# Convert character vector to list of symbols
dots <- lapply(grp_cols, as.symbol)
# Perform frequency counts
df %>%
group_by_(.dots=dots) %>%
summarise(n = n())
output:
Source: local data frame [9 x 3]
Groups: asihckhdoydk
asihckhdoydk a30mvxigxkgh n
1 A A 10
2 A B 10
3 A C 13
4 B A 14
5 B B 10
6 B C 12
7 C A 9
8 C B 12
9 C C 10
Since this question was posted, dplyr added scoped versions of group_by (documentation here). This lets you use the same functions you would use with select, like so:
data = data.frame(
asihckhdoydkhxiydfgfTgdsx = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkghc5cdsvxvyv0ja = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
# get the columns we want to average within
columns = names(data)[-3]
library(dplyr)
df1 <- data %>%
group_by_at(vars(one_of(columns))) %>%
summarize(Value = mean(value))
#compare plyr for reference
df2 <- plyr::ddply(data, columns, plyr::summarize, value=mean(value))
table(df1 == df2, useNA = 'ifany')
## TRUE
## 27
The output from your example question is as expected (see comparison to plyr above and output below):
# A tibble: 9 x 3
# Groups: asihckhdoydkhxiydfgfTgdsx [?]
asihckhdoydkhxiydfgfTgdsx a30mvxigxkghc5cdsvxvyv0ja Value
<fctr> <fctr> <dbl>
1 A A 0.04095002
2 A B 0.24943935
3 A C -0.25783892
4 B A 0.15161805
5 B B 0.27189974
6 B C 0.20858897
7 C A 0.19502221
8 C B 0.56837548
9 C C -0.22682998
Note that since dplyr::summarize only strips off one layer of grouping at a time, you've still got some grouping going on in the resultant tibble (which can sometime catch people by suprise later down the line). If you want to be absolutely safe from unexpected grouping behavior, you can always add %>% ungroup to your pipeline after you summarize.
The support for this in dplyr is currently pretty weak, eventually I think the syntax will be something like:
df %.% group_by(.groups = c("asdfgfTgdsx", "asdfk30v0ja"))
But that probably won't be there for a while (because I need to think through all the consequences).
In the meantime, you can use regroup(), which takes a list of symbols:
library(dplyr)
df <- data.frame(
asihckhdoydk = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkgh = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
df %.%
regroup(list(quote(asihckhdoydk), quote(a30mvxigxkgh))) %.%
summarise(n = n())
If you have have a character vector of column names, you can convert them to the right structure with lapply() and as.symbol():
vars <- setdiff(names(df), "value")
vars2 <- lapply(vars, as.symbol)
df %.% regroup(vars2) %.% summarise(n = n())
String specification of columns in dplyr are now supported through variants of the dplyr functions with names finishing in an underscore. For example, corresponding to the group_by function there is a group_by_ function that may take string arguments. This vignette describes the syntax of these functions in detail.
The following snippet cleanly solves the problem that #sharoz originally posed (note the need to write out the .dots argument):
# Given data and columns from the OP
data %>%
group_by_(.dots = columns) %>%
summarise(Value = mean(value))
(Note that dplyr now uses the %>% operator, and %.% is deprecated).
Update with across() from dplyr 1.0.0
All the answers above are still working, and the solutions with the .dots argument are intruiging.
BUT if you look for a solution that is easier to remember, the new across() comes in handy. It was published 2020-04-03 by Hadley Wickham and can be used in mutate() and summarise() and replace the scoped variants like _at or _all. Above all, it replaces very elegantly the cumbersome non-standard evaluation (NSE) with quoting/unquoting such as !!! rlang::syms().
So the solution with across looks very readable:
data %>%
group_by(across(all_of(columns))) %>%
summarize(Value = mean(value))
Until dplyr has full support for string arguments, perhaps this gist is useful:
https://gist.github.com/skranz/9681509
It contains bunch of wrapper functions like s_group_by, s_mutate, s_filter, etc that use string arguments. You can mix them with the normal dplyr functions. For example
cols = c("cyl","gear")
mtcars %.%
s_group_by(cols) %.%
s_summarise("avdisp=mean(disp), max(disp)") %.%
arrange(avdisp)
It works if you pass it the objects (well, you aren't, but...) rather than as a character vector:
df %.%
group_by(asdfgfTgdsx, asdfk30v0ja) %.%
summarise(Value = mean(value))
> df %.%
+ group_by(asdfgfTgdsx, asdfk30v0ja) %.%
+ summarise(Value = mean(value))
Source: local data frame [9 x 3]
Groups: asdfgfTgdsx
asdfgfTgdsx asdfk30v0ja Value
1 A C 0.046538002
2 C B -0.286359899
3 B A -0.305159419
4 C A -0.004741504
5 B B 0.520126476
6 C C 0.086805492
7 B C -0.052613078
8 A A 0.368410146
9 A B 0.088462212
where df was your data.
?group_by says:
...: variables to group by. All tbls accept variable names, some
will also accept functons of variables. Duplicated groups
will be silently dropped.
which I interpret to mean not the character versions of the names, but how you would refer to them in foo$bar; bar is not quoted here. Or how you'd refer to variables in a formula: foo ~ bar.
#Arun also mentions that you can do:
df %.%
group_by("asdfgfTgdsx", "asdfk30v0ja") %.%
summarise(Value = mean(value))
But you can't pass in something that unevaluated is not a name of a variable in the data object.
I presume this is due to the internal methods Hadley is using to look up the things you pass in via the ... argument.
data = data.frame(
my.a = sample(LETTERS[1:3], 100, replace=TRUE),
my.b = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
group_by(data,newcol=paste(my.a,my.b,sep="_")) %>% summarise(Value=mean(value))
One (tiny) case that is missing from the answers here, that I wanted to make explicit, is when the variables to group by are generated dynamically midstream in a pipeline:
library(wakefield)
df_foo = r_series(rnorm, 10, 1000)
df_foo %>%
# 1. create quantized versions of base variables
mutate_each(
funs(Quantized = . > 0)
) %>%
# 2. group_by the indicator variables
group_by_(
.dots = grep("Quantized", names(.), value = TRUE)
) %>%
# 3. summarize the base variables
summarize_each(
funs(sum(., na.rm = TRUE)), contains("X_")
)
This basically shows how to use grep in conjunction with group_by_(.dots = ...) to achieve this.
General example on using the .dots argument as character vector input to the dplyr::group_by function :
iris %>%
group_by(.dots ="Species") %>%
summarise(meanpetallength = mean(Petal.Length))
Or without a hard coded name for the grouping variable (as asked by the OP):
iris %>%
group_by(.dots = names(iris)[5]) %>%
summarise_at("Petal.Length", mean)
With the example of the OP:
data %>%
group_by(.dots =names(data)[-3]) %>%
summarise_at("value", mean)
See also the dplyr vignette on programming which explains pronouns, quasiquotation, quosures, and tidyeval.
I'm trying to transfer my understanding of plyr into dplyr, but I can't figure out how to group by multiple columns.
# make data with weird column names that can't be hard coded
data = data.frame(
asihckhdoydkhxiydfgfTgdsx = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkghc5cdsvxvyv0ja = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
# get the columns we want to average within
columns = names(data)[-3]
# plyr - works
ddply(data, columns, summarize, value=mean(value))
# dplyr - raises error
data %.%
group_by(columns) %.%
summarise(Value = mean(value))
#> Error in eval(expr, envir, enclos) : index out of bounds
What am I missing to translate the plyr example into a dplyr-esque syntax?
Edit 2017: Dplyr has been updated, so a simpler solution is available. See the currently selected answer.
Just so as to write the code in full, here's an update on Hadley's answer with the new syntax:
library(dplyr)
df <- data.frame(
asihckhdoydk = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkgh = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
# Columns you want to group by
grp_cols <- names(df)[-3]
# Convert character vector to list of symbols
dots <- lapply(grp_cols, as.symbol)
# Perform frequency counts
df %>%
group_by_(.dots=dots) %>%
summarise(n = n())
output:
Source: local data frame [9 x 3]
Groups: asihckhdoydk
asihckhdoydk a30mvxigxkgh n
1 A A 10
2 A B 10
3 A C 13
4 B A 14
5 B B 10
6 B C 12
7 C A 9
8 C B 12
9 C C 10
Since this question was posted, dplyr added scoped versions of group_by (documentation here). This lets you use the same functions you would use with select, like so:
data = data.frame(
asihckhdoydkhxiydfgfTgdsx = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkghc5cdsvxvyv0ja = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
# get the columns we want to average within
columns = names(data)[-3]
library(dplyr)
df1 <- data %>%
group_by_at(vars(one_of(columns))) %>%
summarize(Value = mean(value))
#compare plyr for reference
df2 <- plyr::ddply(data, columns, plyr::summarize, value=mean(value))
table(df1 == df2, useNA = 'ifany')
## TRUE
## 27
The output from your example question is as expected (see comparison to plyr above and output below):
# A tibble: 9 x 3
# Groups: asihckhdoydkhxiydfgfTgdsx [?]
asihckhdoydkhxiydfgfTgdsx a30mvxigxkghc5cdsvxvyv0ja Value
<fctr> <fctr> <dbl>
1 A A 0.04095002
2 A B 0.24943935
3 A C -0.25783892
4 B A 0.15161805
5 B B 0.27189974
6 B C 0.20858897
7 C A 0.19502221
8 C B 0.56837548
9 C C -0.22682998
Note that since dplyr::summarize only strips off one layer of grouping at a time, you've still got some grouping going on in the resultant tibble (which can sometime catch people by suprise later down the line). If you want to be absolutely safe from unexpected grouping behavior, you can always add %>% ungroup to your pipeline after you summarize.
The support for this in dplyr is currently pretty weak, eventually I think the syntax will be something like:
df %.% group_by(.groups = c("asdfgfTgdsx", "asdfk30v0ja"))
But that probably won't be there for a while (because I need to think through all the consequences).
In the meantime, you can use regroup(), which takes a list of symbols:
library(dplyr)
df <- data.frame(
asihckhdoydk = sample(LETTERS[1:3], 100, replace=TRUE),
a30mvxigxkgh = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
df %.%
regroup(list(quote(asihckhdoydk), quote(a30mvxigxkgh))) %.%
summarise(n = n())
If you have have a character vector of column names, you can convert them to the right structure with lapply() and as.symbol():
vars <- setdiff(names(df), "value")
vars2 <- lapply(vars, as.symbol)
df %.% regroup(vars2) %.% summarise(n = n())
String specification of columns in dplyr are now supported through variants of the dplyr functions with names finishing in an underscore. For example, corresponding to the group_by function there is a group_by_ function that may take string arguments. This vignette describes the syntax of these functions in detail.
The following snippet cleanly solves the problem that #sharoz originally posed (note the need to write out the .dots argument):
# Given data and columns from the OP
data %>%
group_by_(.dots = columns) %>%
summarise(Value = mean(value))
(Note that dplyr now uses the %>% operator, and %.% is deprecated).
Update with across() from dplyr 1.0.0
All the answers above are still working, and the solutions with the .dots argument are intruiging.
BUT if you look for a solution that is easier to remember, the new across() comes in handy. It was published 2020-04-03 by Hadley Wickham and can be used in mutate() and summarise() and replace the scoped variants like _at or _all. Above all, it replaces very elegantly the cumbersome non-standard evaluation (NSE) with quoting/unquoting such as !!! rlang::syms().
So the solution with across looks very readable:
data %>%
group_by(across(all_of(columns))) %>%
summarize(Value = mean(value))
Until dplyr has full support for string arguments, perhaps this gist is useful:
https://gist.github.com/skranz/9681509
It contains bunch of wrapper functions like s_group_by, s_mutate, s_filter, etc that use string arguments. You can mix them with the normal dplyr functions. For example
cols = c("cyl","gear")
mtcars %.%
s_group_by(cols) %.%
s_summarise("avdisp=mean(disp), max(disp)") %.%
arrange(avdisp)
It works if you pass it the objects (well, you aren't, but...) rather than as a character vector:
df %.%
group_by(asdfgfTgdsx, asdfk30v0ja) %.%
summarise(Value = mean(value))
> df %.%
+ group_by(asdfgfTgdsx, asdfk30v0ja) %.%
+ summarise(Value = mean(value))
Source: local data frame [9 x 3]
Groups: asdfgfTgdsx
asdfgfTgdsx asdfk30v0ja Value
1 A C 0.046538002
2 C B -0.286359899
3 B A -0.305159419
4 C A -0.004741504
5 B B 0.520126476
6 C C 0.086805492
7 B C -0.052613078
8 A A 0.368410146
9 A B 0.088462212
where df was your data.
?group_by says:
...: variables to group by. All tbls accept variable names, some
will also accept functons of variables. Duplicated groups
will be silently dropped.
which I interpret to mean not the character versions of the names, but how you would refer to them in foo$bar; bar is not quoted here. Or how you'd refer to variables in a formula: foo ~ bar.
#Arun also mentions that you can do:
df %.%
group_by("asdfgfTgdsx", "asdfk30v0ja") %.%
summarise(Value = mean(value))
But you can't pass in something that unevaluated is not a name of a variable in the data object.
I presume this is due to the internal methods Hadley is using to look up the things you pass in via the ... argument.
data = data.frame(
my.a = sample(LETTERS[1:3], 100, replace=TRUE),
my.b = sample(LETTERS[1:3], 100, replace=TRUE),
value = rnorm(100)
)
group_by(data,newcol=paste(my.a,my.b,sep="_")) %>% summarise(Value=mean(value))
One (tiny) case that is missing from the answers here, that I wanted to make explicit, is when the variables to group by are generated dynamically midstream in a pipeline:
library(wakefield)
df_foo = r_series(rnorm, 10, 1000)
df_foo %>%
# 1. create quantized versions of base variables
mutate_each(
funs(Quantized = . > 0)
) %>%
# 2. group_by the indicator variables
group_by_(
.dots = grep("Quantized", names(.), value = TRUE)
) %>%
# 3. summarize the base variables
summarize_each(
funs(sum(., na.rm = TRUE)), contains("X_")
)
This basically shows how to use grep in conjunction with group_by_(.dots = ...) to achieve this.
General example on using the .dots argument as character vector input to the dplyr::group_by function :
iris %>%
group_by(.dots ="Species") %>%
summarise(meanpetallength = mean(Petal.Length))
Or without a hard coded name for the grouping variable (as asked by the OP):
iris %>%
group_by(.dots = names(iris)[5]) %>%
summarise_at("Petal.Length", mean)
With the example of the OP:
data %>%
group_by(.dots =names(data)[-3]) %>%
summarise_at("value", mean)
See also the dplyr vignette on programming which explains pronouns, quasiquotation, quosures, and tidyeval.