I have a data.frame like this -
set.seed(123)
df = data.frame(x=sample(0:1,10,replace=T),y=sample(0:1,10,replace=T),z=1:10)
> df
x y z
1 0 1 1
2 1 0 2
3 0 1 3
4 1 1 4
5 1 0 5
6 0 1 6
7 1 0 7
8 1 0 8
9 1 0 9
10 0 1 10
I would like to remove duplicate rows based on first two columns. Expected output -
df[!duplicated(df[,1:2]),]
x y z
1 0 1 1
2 1 0 2
4 1 1 4
I am specifically looking for a solution using dplyr package.
Here is a solution using dplyr >= 0.5.
library(dplyr)
set.seed(123)
df <- data.frame(
x = sample(0:1, 10, replace = T),
y = sample(0:1, 10, replace = T),
z = 1:10
)
> df %>% distinct(x, y, .keep_all = TRUE)
x y z
1 0 1 1
2 1 0 2
3 1 1 4
Note: dplyr now contains the distinct function for this purpose.
Original answer below:
library(dplyr)
set.seed(123)
df <- data.frame(
x = sample(0:1, 10, replace = T),
y = sample(0:1, 10, replace = T),
z = 1:10
)
One approach would be to group, and then only keep the first row:
df %>% group_by(x, y) %>% filter(row_number(z) == 1)
## Source: local data frame [3 x 3]
## Groups: x, y
##
## x y z
## 1 0 1 1
## 2 1 0 2
## 3 1 1 4
(In dplyr 0.2 you won't need the dummy z variable and will just be
able to write row_number() == 1)
I've also been thinking about adding a slice() function that would
work like:
df %>% group_by(x, y) %>% slice(from = 1, to = 1)
Or maybe a variation of unique() that would let you select which
variables to use:
df %>% unique(x, y)
For completeness’ sake, the following also works:
df %>% group_by(x) %>% filter (! duplicated(y))
However, I prefer the solution using distinct, and I suspect it’s faster, too.
Most of the time, the best solution is using distinct() from dplyr, as has already been suggested.
However, here's another approach that uses the slice() function from dplyr.
# Generate fake data for the example
library(dplyr)
set.seed(123)
df <- data.frame(
x = sample(0:1, 10, replace = T),
y = sample(0:1, 10, replace = T),
z = 1:10
)
# In each group of rows formed by combinations of x and y
# retain only the first row
df %>%
group_by(x, y) %>%
slice(1)
Difference from using the distinct() function
The advantage of this solution is that it makes it explicit which rows are retained from the original dataframe, and it can pair nicely with the arrange() function.
Let's say you had customer sales data and you wanted to retain one record per customer, and you want that record to be the one from their latest purchase. Then you could write:
customer_purchase_data %>%
arrange(desc(Purchase_Date)) %>%
group_by(Customer_ID) %>%
slice(1)
When selecting columns in R for a reduced data-set you can often end up with duplicates.
These two lines give the same result. Each outputs a unique data-set with two selected columns only:
distinct(mtcars, cyl, hp);
summarise(group_by(mtcars, cyl, hp));
If you want to find the rows that are duplicated you can use find_duplicates from hablar:
library(dplyr)
library(hablar)
df <- tibble(a = c(1, 2, 2, 4),
b = c(5, 2, 2, 8))
df %>% find_duplicates()
Related
I am attempting to write some R code that assesses whether or not two dataframes have any matches in their columns. If there are matches, one of the columns in the second dataframe should assign a "link" (via the links variable) to the first dataframe using the id column of the first dataframe.
In the event that there are multiple matches, I am trying to get the "link" variable to randomly select one of the matching id's.
Some reproducible code:
library(dplyr)
df1 = data.frame(ids = c(1:5),
var = c("a","a","c","b","b"))
df2 = data.frame(var = c('c','a','b','b','d'),
links = 0)
Ideally, I would like a resulting dataframe that looks like:
var links
1 c 3
2 a 1 or 2
3 b 4 or 5
4 b 4 or 5
5 d 0
where observations in the links column randomly select ids from df1 when df1$var matches df2$var. In the dataframe above, this is denoted by "or".
Note 1: The links column should be a numeric, I only made it character to allow to write the word "or".
Note 2: If there is not a match between df1$var and df2$var, the links column should remain a 0.
So far, I've gone this route, but I'm unsure about what to put after the ~
linked_df = df2 %>%
mutate(links=case_when(links==0 & var %in% df1$var ~
sample(c(df1$ids),n(),replace=T) # unsure about this line
TRUE ~ links)
I think this is what you want. I've left the ids column in the result, but
it can be removed when the sampling is complete.
library(dplyr)
library(tidyr)
df1_nest = df1 %>%
group_by(var) %>%
summarize(ids = list(ids))
safe_sample = function(x, ...) {
if(length(x) == 1) return(x)
sample(x, ...)
}
set.seed(47)
df2 %>%
left_join(df1_nest) %>%
mutate(
links = sapply(ids, \(x) if(is.null(x)) 0L else safe_sample(x, size = 1))
)
# Joining, by = "var"
# var links ids
# 1 c 3 3
# 2 a 1 1, 2
# 3 b 4 4, 5
# 4 b 5 4, 5
# 5 d 0 NULL
Something like this could do the trick, just a map of a filter of the first dataframe:
df2 %>%
as_tibble() %>%
mutate(links = map(var, ~sample(filter(df1, var == .)$ids), 1),
index = row_number()) %>%
unnest(links, keep_empty = TRUE) %>%
group_by(index) %>%
slice_sample(n = 1) %>%
ungroup() %>%
select(-index)
# # A tibble: 5 × 2
# var links
# <chr> <int>
# 1 c 1
# 2 a 1
# 3 b 4
# 4 b 5
# 5 d NA
Missing something small here and struggling to pass columns to function. I just want to map (or lapply) over columns and perform a custom function on each of the columns. Minimal example here:
library(tidyverse)
set.seed(10)
df <- data.frame(id = c(1,1,1,2,3,3,3,3),
r_r1 = sample(c(0,1), 8, replace = T),
r_r2 = sample(c(0,1), 8, replace = T),
r_r3 = sample(c(0,1), 8, replace = T))
df
# id r_r1 r_r2 r_r3
# 1 1 0 0 1
# 2 1 0 0 1
# 3 1 1 0 1
# 4 2 1 1 0
# 5 3 1 0 0
# 6 3 0 0 1
# 7 3 1 1 1
# 8 3 1 0 0
a function just to filter and counts unique ids remaining in the dataset:
cnt_un <- function(var) {
df %>%
filter({{var}} == 1) %>%
group_by({{var}}) %>%
summarise(n_uniq = n_distinct(id)) %>%
ungroup()
}
it works outside of map
cnt_un(r_r1)
# A tibble: 1 x 2
r_r1 n_uniq
<dbl> <int>
1 1 3
I want to apply the function over all r_r columns to get something like:
df2
# y n_uniq
# 1 r_r1 3
# 2 r_r2 2
# 3 r_r3 2
I thought the following would work but doesnt
map(dplyr::select(df, matches("r_r")), ~ cnt_un(.x))
any suggestions? thanks
I'm not sure if there's a direct tidyeval way to do this with something like map. The issue you're running into is that in calling map(df, *whatever_function*), the function is being called on each column of df as a vector, whereas your function expects a bare column name in the tidyeval style. To verify that:
map(df, class)
will return "numeric" for each column.
An alternative is to iterate over column names as strings, and convert those to symbols; this takes just one additional line in the function.
library(dplyr)
library(tidyr)
library(purrr)
cnt_un_name <- function(varname) {
var <- ensym(varname)
df %>%
filter({{var}} == 1) %>%
group_by({{var}}) %>%
summarise(n_uniq = n_distinct(id)) %>%
ungroup()
}
Calling the function is a little awkward because it keeps only the relevant column names (calling on "r_r1" gets columns "r_r1" and "n_uniq", etc). One way is to get the vector of column names you want, name it so you can add an ID column in map_dfr, and drop the extra columns, since they'll be mostly NA.
grep("^r_r\\d+", names(df), value = TRUE) %>%
set_names() %>%
map_dfr(cnt_un_name, .id = "y") %>%
select(y, n_uniq)
#> # A tibble: 3 x 2
#> y n_uniq
#> <chr> <int>
#> 1 r_r1 3
#> 2 r_r2 2
#> 3 r_r3 2
A better way is to call the function, then bind after reshaping.
grep("^r_r\\d+", names(df), value = TRUE) %>%
map(cnt_un_name) %>%
map_dfr(pivot_longer, 1, names_to = "y") %>%
select(y, n_uniq)
# same output as above
Alternatively (and maybe better/more scaleable) would be to do the column renaming inside the function definition.
Here's a base R solution that uses lapply. The tricky bit is that your function isn't actually running on single columns; it's using id, too, so you can't use canned functions that iterate column-wise.
do.call(rbind, lapply(grep("r_r", colnames(df), value = TRUE), function(i) {
X <- subset(df, df[,i] == 1)
row <- data.frame(y = i, n_uniq = length(unique(X$id)), stringsAsFactors = FALSE)
}))
y n_uniq
1 r_r1 2
2 r_r2 3
3 r_r3 2
Here is another solution. I changed the syntax of your function. Now you supply the pattern of the columns you want to select.
cnt_un <- function(var_pattern) {
df %>%
pivot_longer(cols = contains(var_pattern), values_to = "vals", names_to = "y") %>%
filter(vals == 1) %>%
group_by(y) %>%
summarise(n_uniq = n_distinct(id)) %>%
ungroup()
}
cnt_un("r_r")
#> # A tibble: 3 x 2
#> y n_uniq
#> <chr> <int>
#> 1 r_r1 2
#> 2 r_r2 3
#> 3 r_r3 2
I am handling a large dataset. First, for certain columns (X1, X2, ...), I am trying to identify a range of value (a, b) consists of repeated value (a > n, b > n). Next, I wish to filter row based on the condition which matches respective columns to result given in the previous step.
Here is a reproducible example simulating the scenario I am facing,
library(tidyverse)
set.seed(1122)
vecs <- lapply(X = 1:2, function(x) rep(c(1, 2, 3), times = 10) %>% sample() %>% head(10))
names(vecs) <- paste0("col_", 1:2)
dat <- vecs %>% as.data.frame()
dat
col_1 col_2
1 3 2
2 1 1
3 1 1
4 1 2
5 1 2
6 3 3
7 3 3
8 2 1
9 1 3
10 2 2
I am able to identify the range by the following method,
# Which col has repeated value more than 3 appearances?
more_than_3 <- function(df, var){
var <- rlang::sym(var)
df %>%
group_by(!!var) %>%
summarise(n = n()) %>%
filter(n > 3) %>%
pull(!!var) %>%
range()
}
cols_name <- c("col_1", "col_2")
some_range <- purrr::map(cols_name, more_than_3, df = dat)
names(some_range) <- cols_name
some_range
$col_1
[1] 1 1
$col_2
[1] 2 2
However, to filter out values that fall outside the upper limit, this is what I do.
dat %>%
filter(col_1 <= some_range[["col_1"]][2],
col_2 <= some_range[["col_2"]][2])
col_1 col_2
1 1 1
2 1 1
3 1 2
4 1 2
I believe there must be a more efficient and elegant way of filtering the result based on tidy evaluation. Can someone point me to the right direction?
Many thanks in advance.
First let's try to create a small function that creates a single filter expression for one column. This function will take a symbol and then transform to string but it could be the other way around:
new_my_filter_call_upper <- function(sym, range) {
col_name <- as.character(sym)
col_range <- range[[col_name]]
if (is.null(col_range)) {
stop(sprintf("Can't find column `%s` to compute range", col_name), call. = FALSE)
}
expr(!!sym < !!col_range[[2]])
}
Let's try it:
new_my_filter_call_upper(quote(foobar), some_range)
#> Error: Can't find column `foobar` to compute range
# It works!
new_my_filter_call_upper(quote(col_1), some_range)
#> col_1 < 3
Now we're ready to create a pipeline verbs that will take a data frame and bare column names.
# Probably cleaner to pass range as argument. Prefix with dot to allow
# columns named `range`.
my_filter <- function(.data, ..., .range) {
# ensyms() guarantees there won't be complex expressions
syms <- rlang::ensyms(...)
# Now let's map the function to create many calls:
calls <- purrr::map(syms, new_my_filter_call_upper, range = .range)
# And we're ready to filter with those expressions:
dplyr::filter(.data, !!!calls)
}
Let's try it:
dat %>% my_filter(col_1, col_2, .range = some_range)
#> col_1 col_2 NA.
#> 1 2 1 1
#> 2 2 2 1
We could use map2
library(purrr)
map2(dat, some_range, ~ .x < .y[2]) %>%
reduce(`&`) %>%
dat[.,]
# col_1 col_2
#1 2 2
#2 1 1
#3 1 2
#6 1 1
Or with pmap
pmap(list(dat, some_range %>%
map(2)), `<`) %>%
reduce(`&`) %>%
dat[.,]
I have a dataframe with groups that essentially looks like this
DF <- data.frame(state = c(rep("A", 3), rep("B",2), rep("A",2)))
DF
state
1 A
2 A
3 A
4 B
5 B
6 A
7 A
My question is how to count the number of consecutive rows where the first value is repeated in its first "block". So for DF above, the result should be 3. The first value can appear any number of times, with other values in between, or it may be the only value appearing.
The following naive attempt fails in general, as it counts all occurrences of the first value.
DF %>% mutate(is_first = as.integer(state == first(state))) %>%
summarize(count = sum(is_first))
The result in this case is 5. So, hints on a (preferably) dplyr solution to this would be appreciated.
You can try:
rle(as.character(DF$state))$lengths[1]
[1] 3
In your dplyr chain that would just be:
DF %>% summarize(count_first = rle(as.character(state))$lengths[1])
# count_first
# 1 3
Or to be overzealous with piping, using dplyr and magrittr:
library(dplyr)
library(magrittr)
DF %>% summarize(count_first = state %>%
as.character %>%
rle %$%
lengths %>%
first)
# count_first
# 1 3
Works also for grouped data:
DF <- data.frame(group = c(rep(1,4),rep(2,3)),state = c(rep("A", 3), rep("B",2), rep("A",2)))
# group state
# 1 1 A
# 2 1 A
# 3 1 A
# 4 1 B
# 5 2 B
# 6 2 A
# 7 2 A
DF %>% group_by(group) %>% summarize(count_first = rle(as.character(state))$lengths[1])
# # A tibble: 2 x 2
# group count_first
# <dbl> <int>
# 1 1 3
# 2 2 1
No need of dplyrhere but you can modify this example to use it with dplyr. The key is the function rle
state = c(rep("A", 3), rep("B",2), rep("A",2))
x = rle(state)
DF = data.frame(len = x$lengths, state = x$values)
DF
# get the longest run of consecutive "A"
max(DF[DF$state == "A",]$len)
I have data looking somewhat similar to this:
number type results
1 5 x, y, z
2 6 a
3 8 x
1 5 x, y
Basically, I have data in Excel that has commas in a couple of individual cells and I need to count each value that is separated by a comma, after a certain requirement is met by subsetting.
Question: How do I go about receiving the sum of 5 when subsetting the data with number == 1 and type == 5, in R?
If we need the total count, then another option is str_count after subsetting
library(stringr)
with(df, sum(str_count(results[number==1 & type==5], "[a-z]"), na.rm = TRUE))
#[1] 5
Or with gregexpr from base R
with(df, sum(lengths(gregexpr("[a-z]", results[number==1 & type==5])), na.rm = TRUE))
#[1] 5
If there are no matching pattern for an element, use
with(df, sum(unlist(lapply(gregexpr("[a-z]",
results[number==1 & type==5]), `>`, 0)), na.rm = TRUE))
Here is an option using dplyr and tidyr. filter function can filter the rows based on conditions. separate_rows can separate the comma. group_by is to group the data. tally can count the numbers.
dt2 <- dt %>%
filter(number == 1, type == 5) %>%
separate_rows(results) %>%
group_by(results) %>%
tally()
# # A tibble: 3 x 2
# results n
# <chr> <int>
# 1 x 2
# 2 y 2
# 3 z 1
Or you can use count(results) only as the following code shows.
dt2 <- dt %>%
filter(number == 1, type == 5) %>%
separate_rows(results) %>%
count(results)
DATA
dt <- read.table(text = "number type results
1 5 'x, y, z'
2 6 a
3 8 x
1 5 'x, y'",
header = TRUE, stringsAsFactors = FALSE)
Here is a method using base R. You split results on the commas and get the length of each list, then add these up grouping by number.
aggregate(sapply(strsplit(df$results, ","), length), list(df$number), sum)
Group.1 x
1 1 5
2 2 1
3 3 1
Your data:
df = read.table(text="number type results
1 5 'x, y, z'
2 6 'a'
3 8 'x'
1 5 'x, y'",
header=TRUE, stringsAsFactors=FALSE)