I have a dataframe that consists of 3 different columns (a,b and noise).
I want to apply a function on all distinct combinations of the two first columns together with the mean of the third variable and save it in a new column named c.
My first thought was to solve it with the following code
library(dplyr)
df <- data.frame(a = rep(c(1,2,3),each=9),
b = rep(c(1,2,3),length.out=3*9),
noise = rnorm(9*3*1000))
f <- function(a,b,c) a + b + c
result <- df %>% group_by(a,b) %>% summarise(c = f(a,b,mean(noise)))
To my surprise this gave the error "Error: expecting a single value".
So dlyr still treats a and b as vectors.
So the problem could then be solved with the somewhat confusing code
result <- df %>% group_by(a,b) %>% summarise(c = f(a[1],b[1],mean(noise)))
My questions are:
Why do dplyr keep grouped variables as vectors (are there any benefits of this?)
Is there any better way to solve this problem using dplyr?
Related
As a follow up to this question, I'm using dplyr's group_split() to make dataframes / tibbles based on a levels of a column. Continuing off of this question, I want to split off of two columns instead of 1. When I try to split and name the columns, it attributes the wrong names to some of the datasets.
Here's a simple example:
library(dplyr)
#Sample dataset to intuitively illustrate issue
example <- tibble(number = c(1:6),
even_or_odd = c("odd", "even", "odd", "even", "odd", "even"),
prime_or_not = c("prime", "prime", "prime", "not", "prime", "not")) %>%
mutate(type = paste0(even_or_odd, "_", prime_or_not)) %>%
mutate(type_factor = factor(type, levels = unique(type)))
#Does group split to make 3 datasets
the_test <- example %>%
group_split(even_or_odd, prime_or_not) %>%
setNames(unique(example$type_factor))
#The data sets with some being correct but others not
even_prime <- the_test["even_prime"]$even_prime #works!
even_not <- the_test["even_not"]$even_not #wrong label :`-(
odd_prime <- the_test["odd_prime"]$odd_prime #wrong label :`-(
odd_not <- the_test["odd_not"]$odd_not #works--correctly throws an error!
My question: how do I ensure that my group names will be attributed to the right dataset and avoid the issues here with even_not and odd_prime being mixed up?
In my actual dataset, I have 50+ combinations, so typing them all out manually is not an option. In addition, my actual dataset will have some combinations that don't consistently exist (like the (like the odd not prime combination here), so relying on index isn't an option.
Instead of splitting by the two columns, use the factor column that was created, which ensures that it splits by the order of the levels created in the type_factor. In addition, using the unique on type_factor can have some issues if the order of the values in 'type_factor' is different i.e. unique gets the first non-duplicated value based on its occurrence. Instead, levels is better. In fact, it may be more appropriate to droplevels as well in case of unused levels.
the_test <- example %>%
group_split(type_factor) %>%
setNames(levels(example$type_factor))
group_split returns unnamed list. If we want to avoid the pain of renaming incorrectly, use split from base R which does return a named list. Thus, it can return in any order as long as the key/value pairs are correct
# 1 - return in a different order based on alphabetic order
split(example, example[c("even_or_odd", "prime_or_not")], drop = TRUE)
# 2 - return order based on the levels of the factor column
split(example, example$type_factor)
# 3 - With dplyr pipe
example %>%
split(.$type_factor)
# 4 - or using magrittr exposition operator
library(magrittr)
example %$%
split(x = ., f = type_factor)
Oh, of course the moment I post it, I realize that an easy solution existed:
Just change the group split to the new variable and it works!
library(dplyr)
#Does group split to make 3 datasets
the_test <- example %>%
group_split(type_factor) %>%
setNames(unique(example$type_factor))
#The data sets with some being correct but others not
even_prime <- the_test["even_prime"]$even_prime #works!
even_not <- the_test["even_not"]$even_not #works now!
odd_prime <- the_test["odd_prime"]$odd_prime #works now!
odd_not <- the_test["odd_not"]$odd_not #works--correctly throws an error!
I have a dataframe ("df") with a number of columns that I would like to estimate the weighted means of, weighting by population (df$Population), and grouping by commuting zone (df$cz).
This is the list of columns I would like to estimate the weighted means of:
vlist = c("Public_Welf_Total_Exp", "Welf_Cash_Total_Exp", "Welf_Cash_Cash_Assist", "Welf_Ins_Total_Exp","Total_Educ_Direct_Exp", "Higher_Ed_Total_Exp", "Welf_NEC_Cap_Outlay","Welf_NEC_Direct_Expend", "Welf_NEC_Total_Expend", "Total_Educ_Assist___Sub", "Health_Total_Expend", "Total_Hospital_Total_Exp", "Welf_Vend_Pmts_Medical","Hosp_Other_Total_Exp","Unemp_Comp_Total_Exp", "Unemp_Comp_Cash___Sec", "Total_Unemp_Rev", "Hous___Com_Total_Exp", "Hous___Com_Construct")
This is the code I have been using:
df = df %>% group_by(cz) %>% mutate_at(vlist, weighted.mean(., df$Population))
I have also tried:
df = df %>% group_by(cz) %>% mutate_at(vlist, function(x) weighted.mean(x, df$Population))
As well as tested the following code on only 2 columns:
df = df %>% group_by(cz) %>% mutate_at(vars(Public_Welf_Total_Exp, Welf_Cash_Total_Exp), weighted.mean(., df$Population))
However, everything I have tried gives me the following error, even though there are no NAs in any of my variables:
Error in weighted.mean.default(., df$Population) :
'x' and 'w' must have the same length
I understand that I could do the following estimation using lapply, but I don't know how to group by another variable using lapply. I would appreciate any suggestions!
There is a lot to unpack here...
Probably you mean summarise instead of mutate, because with mutate you would just replicate your result for each row.
mutate_at and summarise_at are subseeded and you should use across instead.
the reason why your code wasn't working was because you did not write your function as a formula (you did not add ~ at the beginning), also you were using df$Population instead of Population. When you write Population, summarise knows you're talking about the column Population which, at that point, is grouped like the rest of the dataframe. When you use df$Population you are calling the column of the original dataframe without grouping. Not only it is wrong, but you would also get an error because the length of the variable you are trying to average and the lengths of the weights provided by df$Population would not correspond.
Here is how you could do it:
library(dplyr)
df %>%
group_by(cz) %>%
summarise(across(vlist, weighted.mean, Population),
.groups = "drop")
If you really need to use summarise_at (and probably you are using an old version of dplyr [lower than 1.0.0]), then you could do:
df %>%
group_by(cz) %>%
summarise_at(vlist, ~weighted.mean(., Population)) %>%
ungroup()
I considered df and vlist like the following:
vlist <- c("Public_Welf_Total_Exp", "Welf_Cash_Total_Exp", "Welf_Cash_Cash_Assist", "Welf_Ins_Total_Exp","Total_Educ_Direct_Exp", "Higher_Ed_Total_Exp", "Welf_NEC_Cap_Outlay","Welf_NEC_Direct_Expend", "Welf_NEC_Total_Expend", "Total_Educ_Assist___Sub", "Health_Total_Expend", "Total_Hospital_Total_Exp", "Welf_Vend_Pmts_Medical","Hosp_Other_Total_Exp","Unemp_Comp_Total_Exp", "Unemp_Comp_Cash___Sec", "Total_Unemp_Rev", "Hous___Com_Total_Exp", "Hous___Com_Construct")
df <- as.data.frame(matrix(rnorm(length(vlist) * 100), ncol = length(vlist)))
names(df) <- vlist
df$cz <- rep(letters[1:10], each = 10)
df$Population <- runif(100)
Some sample data:
df <- data.frame(lang = rep(c("A", "B", "C"), 3),
answer = rep(c("1", "2", "3"), each=3))
I am getting an error when I try to call a variable that I recently grouped by:
df2 <- df %>%
Total = count(lang) %>% # count is short hand for tally + group_by()
filter(answer=='2') %>%
mutate(prop = NROW(answer)/NROW(Total))
Error in group_vars(x) : object 'lang' not found
I would like a new column on my dataframe that says the proportion of the answer '2' to total observations in each level of lang. So how many times does '2' occur in 'A' in proportion to the total number of observations in 'A'?
Here's a solution that does what you want:
df %>%
group_by(lang) %>%
summarize(
prop = length(lang[answer==2])/n()
)
Here, we group by the variable or variables that you want set as the unique groups you want to get the proportion of and then use summarize to calculate the length of the vector of one of the variables where answer is equal to 2 and divide that by the number of rows in the grouping. If, for whatever reason, you want the prop column AND the answer column, just change summarize to mutate.
The reason you were getting the error about not finding lang is because count needs to be used as a function like mutate, i.e.
df %>%
count(lang, name = "Total")
You could achieve the same thing adapting your code, but you should use add_count (so your answer column is preserved) or mutate(Total = n()). However, group_by was designed to address problems such as this and is definitely worth spending some time to learn about.
df %>%
add_count(lang, name = "Total") %>%
filter(answer == 2) %>%
add_count(lang, name = "Twos") %>%
distinct(lang, .keep_all = TRUE) %>%
mutate(prop = Twos/Total) %>%
select(lang, prop)
Alternate solution with data.table
I prefer to use data.table than data frames everywhere personally. Here is the implementation with that method, although admittedly it looks a bit more cryptic than the solution in dplyr (The syntax to accomplish something like this may be more involved, but getting used to it ends up giving you a whole bag of tricks, and with simple queries the syntax actually looks better)
You end up trying to use "lang" like its a variable, when its a name of a column.
To get the values requested, 0.3333 for each,
library(data.table)
df <- data.table(df)
df[, nrow(.SD[answer == 2])/nrow(.SD), by="lang"]
lang V1
1: A 0.3333333
2: B 0.3333333
3: C 0.3333333
(the special variable .SD allows you to manipulate every subset of the data, split by by)
This may well have an answer elsewhere but I'm having trouble formulating the words of the question to find what I need.
I have two dataframes, A and B, with A having many more rows than B. I want to look up a value from B based on a column of A, and add it to another column of A. Something like:
A$ColumnToAdd + B[ColumnToMatch == A$ColumnToMatch,]$ColumnToAdd
But I get, with a load of NAs:
Warning in `==.default`: longer object length is not a multiple of shorter object length
I could do it with a messy for-loop but I'm looking for something faster & elegant.
Thanks
If I understood your question correctly, you're looking for a merge or a join, as suggested in the comments.
Here's a simple example for both using dummy data that should fit what you described.
library(tidyverse)
# Some dummy data
ColumnToAdd <- c(1,1,1,1,1,1,1,1)
ColumnToMatch <- c('a','b','b','b','c','a','c','d')
A <- data.frame(ColumnToAdd, ColumnToMatch)
ColumnToAdd <- c(1,2,3,4)
ColumnToMatch <- c('a','b','c','d')
B <- data.frame(ColumnToAdd, ColumnToMatch)
# Example using merge
A %>%
merge(B, by = c("ColumnToMatch")) %>%
mutate(sum = ColumnToAdd.x + ColumnToAdd.y)
# Example using join
A %>%
inner_join(B, by = c("ColumnToMatch")) %>%
mutate(sum = ColumnToAdd.x + ColumnToAdd.y)
The advantages of the dplyr versions over merge are:
rows are kept in existing order
much faster
tells you what keys you're merging by (if you don't supply)
also work with database tables.
I am trying to transfer from plyr to dplyr. However, I still can't seem to figure out how to call on own functions in a chained dplyr function.
I have a data frame with a factorised ID variable and an order variable. I want to split the frame by the ID, order it by the order variable and add a sequence in a new column.
My plyr functions looks like this:
f <- function(x) cbind(x[order(x$order_variable), ], Experience = 0:(nrow(x)-1))
data <- ddply(data, .(ID_variable), f)
In dplyr I though this should look something like this
f <- function(x) cbind(x[order(x$order_variable), ], Experience = 0:(nrow(x)-1))
data <- data %>% group_by(ID_variable) %>% f
Can anyone tell me how to modify my dplyr call to successfully pass my own function and get the same functionality my plyr function provides?
EDIT: If I use the dplyr formula as described here, it DOES pass an object to f. However, while plyr seems to pass a number of different tables (split by the ID variable), dplyr does not pass one table per group but the ENTIRE table (as some kind of dplyr object where groups are annotated), thus when I cbind the Experience variable it appends a counter from 0 to the length of the entire table instead of the single groups.
I have found a way to get the same functionality in dplyr using this approach:
data <- data %>%
group_by(ID_variable) %>%
arrange(ID_variable,order_variable) %>%
mutate(Experience = 0:(n()-1))
However, I would still be keen to learn how to pass grouped variables split into different tables to own functions in dplyr.
For those who get here from google. Let's say you wrote your own print function.
printFunction <- function(dat) print(dat)
df <- data.frame(a = 1:6, b = 1:2)
As it was asked here
df %>%
group_by(b) %>%
printFunction(.)
prints entire data. To get dplyr print multiple tables grouped by, you should use do
df %>%
group_by(b) %>%
do(printFunction(.))