I'm trying to subset a dataframe in R by checking if each value is present in a specific list and keeping it if it is. For instance in the following dataframe:
x <- data.frame(A = sample(1:5, 5),
B = sample(1:5, 5),
C = sample(1:5, 5))
A B C
1 2 2 1
2 3 3 3
3 1 4 4
4 4 5 2
5 5 1 5
How could I subset it to include only the values 1, 3 and 4, giving the following as a result:
A B C
1 1
2 3 3 3
3 4 4
4 4
5 1
It doesn't matter what happens to the missing values - they could be changed to NA if this is easier. From browsing similar questions it seems that lapply might do it, but as a novice I'm struggling to apply what I've seen to this scenario.
set.seed(47)
x <- data.frame(A = sample(1:5, 5),
B = sample(1:5, 5),
C = sample(1:5, 5))
# with lapply
keep_vals = c(1, 3, 4)
x[] = lapply(x, function(y) {
y[! y %in% keep_vals] = NA
return(y)
})
x
# A B C
# 1 3 1 1
# 2 1 NA NA
# 3 NA NA 4
# 4 4 3 NA
# 5 NA 4 3
Or with a for loop:
set.seed(47) # reset data
x <- data.frame(A = sample(1:5, 5),
B = sample(1:5, 5),
C = sample(1:5, 5))
keep_vals = c(1, 3, 4)
for (i in 1:ncol(x)) {
x[, i][!x[, i] %in% keep_vals] <- NA
}
x
# A B C
# 1 3 1 1
# 2 1 NA NA
# 3 NA NA 4
# 4 4 3 NA
# 5 NA 4 3
With dplyr
x %>% mutate_all(
~replace(., !. %in% keep_vals, NA)
)
# A B C
# 1 3 1 1
# 2 1 NA NA
# 3 NA NA 4
# 4 4 3 NA
# 5 NA 4 3
using dplyr::bind_rows
do.call(bind_rows,apply(x,1, function(a) a[a %in% c(1,3,4)]))
# A tibble: 5 x 3
A B C
<int> <int> <int>
1 4 NA NA
2 1 1 1
3 3 3 NA
4 NA NA 4
5 NA 4 3
Collapsing each row to the matching numbers, and adjusting each length to ncol. (Assuming you want to "left-align" your numbers, as shown in your expected output.)
d <- setNames(as.data.frame(t(apply(d, 1, function(x) {
x <- x[x %in% c(1, 3, 4)]
`length<-`(x, ncol(d))
}))), names(d))
d
# A B C
# 1 1 NA NA
# 2 3 3 3
# 3 1 4 4
# 4 4 NA NA
# 5 NA NA NA
Since apply throws a matrix, we tell R that we want the transpose as.data.frame and setNames to restore those.
Note, that I changed line 5 of your example data so that it doesn't contain any of the matching numbers, so as not to make it too easy.
Data
d <- read.table(text="A B C
1 2 2 1
2 3 3 3
3 1 4 4
4 4 5 2
5 5 2 5", header=TRUE)
Related
Suppose I have this dataframe
df <- data.frame(
x=c(1, NA, NA, 4, 5, NA),
y=c(NA, 2, 3, NA, NA, 6)
which looks like this
x y
1 1 NA
2 NA 2
3 NA 3
4 4 NA
5 5 NA
6 NA 6
How can I merge the two columns into one? Basically the NA values are in complementary rows. It would be nice to also obtain (in the process) a flag column containing 0 if the entry comes from x and 1 if the entry comes from y.
We can try using the coalesce function from the dplyr package:
df$merged <- coalesce(df$x, df$y)
df$flag <- ifelse(is.na(df$y), 0, 1)
df
x y merged flag
1 1 NA 1 0
2 NA 2 2 1
3 NA 3 3 1
4 4 NA 4 0
5 5 NA 5 0
6 NA 6 6 1
We can also use base R methods with max.col on the logical matrix to get the column index, cbind with row index and extract the values that are not NA
df$merged <- df[cbind(seq_len(nrow(df)), max.col(!is.na(df)))]
df$flag <- +(!is.na(df$y))
df
# x y merged flag
#1 1 NA 1 0
#2 NA 2 2 1
#3 NA 3 3 1
#4 4 NA 4 0
#5 5 NA 5 0
#6 NA 6 6 1
Or we can use fcoalesce from data.table which is written in C and is multithreaded for numeric and factor types.
library(data.table)
setDT(df)[, c('merged', 'flag' ) := .(fcoalesce(x, y), +(!is.na(y)))]
df
# x y merged flag
#1: 1 NA 1 0
#2: NA 2 2 1
#3: NA 3 3 1
#4: 4 NA 4 0
#5: 5 NA 5 0
#6: NA 6 6 1
You can do that using dplyr as follows;
library(dplyr)
# Creating dataframe
df <-
data.frame(
x = c(1, NA, NA, 4, 5, NA),
y = c(NA, 2, 3, NA, NA, 6))
df %>%
# If x is null then replace it with y
mutate(merged = coalesce(x, y),
# If x is null then put 1 else put 0
flag = if_else(is.na(x), 1, 0))
# x y merged flag
# 1 NA 1 0
# NA 2 2 1
# NA 3 3 1
# 4 NA 4 0
# 5 NA 5 0
# NA 6 6 1
I want to make value of each row of column A , NA ,where column B is 2:
data
A B
1 2
2 4
NA 5
6 2
output
A B
NA 2
2 4
NA 5
NA 2
first and last row of B was 2 so A got NA in those.
Here's a way using ifelse in base R -
df$A <- ifelse(df$B == 2, NA_real_, df$A)
set.seed(0)
df <- data.frame(A = sample(1:10, size=5, replace=T),
B = sample(1:10, size=5, replace=T))
df
A B
1 9 7
2 4 2
3 7 3
4 1 1
5 2 5
df$A[df$B == 2] <- NA
df
A B
1 9 7
2 NA 2
3 7 3
4 1 1
5 2 5
I am trying to recode NA values to 0 in a subset of columns using the following dataset:
set.seed(1)
df <- data.frame(
id = c(1:10),
trials = sample(1:3, 10, replace = T),
t1 = c(sample(c(1:9, NA), 10)),
t2 = c(sample(c(1:7, rep(NA, 3)), 10)),
t3 = c(sample(c(1:5, rep(NA, 5)), 10))
)
Each row has a certain number of trials associated with it (between 1-3), specified by the trials column. columns t1-t3 represent scores for each trial.
The number of trials indicates the subset of columns in which NAs should be recoded to 0: NAs that are within the number of trials represent missing data, and should be recoded as 0, while NAs outside the number of trials are not meaningful, and should remain NAs. So, for a row where trials == 3, an NA in column t3 would be recoded as 0, but in a row where trials == 2, an NA in t3 would remain an NA.
So, I tried using this function:
replace0 <- function(x, num.sun) {
x[which(is.na(x[1:(num.sun + 2)]))] <- 0
return(x)
}
This works well for single vectors. When I try applying the same function to a data frame with apply(), though:
apply(df, 1, replace0, num.sun = df$trials)
I get a warning saying:
In 1:(num.sun + 2) :
numerical expression has 10 elements: only the first used
The result is that instead of having the value of num.sun change every row according to the value in trials, apply() simply uses the first value in the trials column for every single row. How could I apply the function so that the num.sun argument changes according to the value of df$trials?
Thanks!
Edit: as some have commented, the original example data had some non-NA scores that didn't make sense according to the trials column. Here's a corrected dataset:
df <- data.frame(
id = c(1:5),
trials = c(rep(1, 2), rep(2, 1), rep(3, 2)),
t1 = c(NA, 7, NA, 6, NA),
t2 = c(NA, NA, 3, 7, 12),
t3 = c(NA, NA, NA, 4, NA)
)
Another approach:
# create an index of the NA values
w <- which(is.na(df), arr.ind = TRUE)
# create an index with the max column by row where an NA is allowed to be replaced by a zero
m <- matrix(c(1:nrow(df), (df$trials + 2)), ncol = 2)
# subset 'w' such that only the NA's which fall in the scope of 'm' remain
i <- w[w[,2] <= m[,2][match(w[,1], m[,1])],]
# use 'i' to replace the allowed NA's with a zero
df[i] <- 0
which gives:
> df
id trials t1 t2 t3
1 1 1 3 NA 5
2 2 2 2 2 NA
3 3 2 6 6 4
4 4 3 0 1 2
5 5 1 5 NA NA
6 6 3 7 0 0
7 7 3 8 7 0
8 8 2 4 5 1
9 9 2 1 3 NA
10 10 1 9 4 3
You could easily wrap this in a function:
replace.NA.with.0 <- function(df) {
w <- which(is.na(df), arr.ind = TRUE)
m <- matrix(c(1:nrow(df), (df$trials + 2)), ncol = 2)
i <- w[w[,2] <= m[,2][match(w[,1], m[,1])],]
df[i] <- 0
return(df)
}
Now, using replace.NA.with.0(df) will produce the above result.
As noted by others, some rows (1, 3 & 10) have more values than trails. You could tackle that problem by rewriting the above function to:
replace.with.NA.or.0 <- function(df) {
w <- which(is.na(df), arr.ind = TRUE)
df[w] <- 0
v <- tapply(m[,2], m[,1], FUN = function(x) tail(x:5,-1))
ina <- matrix(as.integer(unlist(stack(v)[2:1])), ncol = 2)
df[ina] <- NA
return(df)
}
Now, using replace.with.NA.or.0(df) produces the following result:
id trials t1 t2 t3
1 1 1 3 NA NA
2 2 2 2 2 NA
3 3 2 6 6 NA
4 4 3 0 1 2
5 5 1 5 NA NA
6 6 3 7 0 0
7 7 3 8 7 0
8 8 2 4 5 NA
9 9 2 1 3 NA
10 10 1 9 NA NA
Here I just rewrite your function using double subsetting x[paste0('t',x['trials'])], which overcome the problem in the other two solutions with row 6
replace0 <- function(x){
#browser()
x_na <- x[paste0('t',x['trials'])]
if(is.na(x_na)){x[paste0('t',x['trials'])] <- 0}
return(x)
}
t(apply(df, 1, replace0))
id trials t1 t2 t3
[1,] 1 1 3 NA 5
[2,] 2 2 2 2 NA
[3,] 3 2 6 6 4
[4,] 4 3 NA 1 2
[5,] 5 1 5 NA NA
[6,] 6 3 7 NA 0
[7,] 7 3 8 7 0
[8,] 8 2 4 5 1
[9,] 9 2 1 3 NA
[10,] 10 1 9 4 3
Here is a way to do it:
x <- is.na(df)
df[x & t(apply(x, 1, cumsum)) > 3 - df$trials] <- 0
The output looks like this:
> df
id trials t1 t2 t3
1 1 1 3 NA 5
2 2 2 2 2 NA
3 3 2 6 6 4
4 4 3 0 1 2
5 5 1 5 NA NA
6 6 3 7 0 0
7 7 3 8 7 0
8 8 2 4 5 1
9 9 2 1 3 NA
10 10 1 9 4 3
> x <- is.na(df)
> df[x & t(apply(x, 1, cumsum)) > 3 - df$trials] <- 0
> df
id trials t1 t2 t3
1 1 1 3 NA 5
2 2 2 2 2 NA
3 3 2 6 6 4
4 4 3 0 1 2
5 5 1 5 NA NA
6 6 3 7 0 0
7 7 3 8 7 0
8 8 2 4 5 1
9 9 2 1 3 NA
10 10 1 9 4 3
Note: row 1/3/10, is problematic since there are more non-NA values than the trials.
Here's a tidyverse way, note that it doesn't give the same output as other solutions.
Your example data shows results for trials that "didn't happen", I assumed your real data doesn't.
library(tidyverse)
df %>%
nest(matches("^t\\d")) %>%
mutate(data = map2(data,trials,~mutate_all(.,replace_na,0) %>% select(.,1:.y))) %>%
unnest
# id trials t1 t2 t3
# 1 1 1 3 NA NA
# 2 2 2 2 2 NA
# 3 3 2 6 6 NA
# 4 4 3 0 1 2
# 5 5 1 5 NA NA
# 6 6 3 7 0 0
# 7 7 3 8 7 0
# 8 8 2 4 5 NA
# 9 9 2 1 3 NA
# 10 10 1 9 NA NA
Using the more commonly used gather strategy this would be:
df %>%
gather(k,v,matches("^t\\d")) %>%
arrange(id) %>%
group_by(id) %>%
slice(1:first(trials)) %>%
mutate_at("v",~replace(.,is.na(.),0)) %>%
spread(k,v)
# # A tibble: 10 x 5
# # Groups: id [10]
# id trials t1 t2 t3
# <int> <int> <dbl> <dbl> <dbl>
# 1 1 1 3 NA NA
# 2 2 2 2 2 NA
# 3 3 2 6 6 NA
# 4 4 3 0 1 2
# 5 5 1 5 NA NA
# 6 6 3 7 0 0
# 7 7 3 8 7 0
# 8 8 2 4 5 NA
# 9 9 2 1 3 NA
# 10 10 1 9 NA NA
I have 2 data frames with different rownames, e.g.:
df1 <- data.frame(A = c(1,3,7,1,5), B = c(5,2,9,5,5), C = c(1,1,3,4,5))
df2 <- data.frame(A = c(4,3,2), B = c(4,4,9), C = c(3,9,3))
rownames(df2) <- c(1, 3, 6)
> df1
A B C
1 1 5 1
2 3 2 1
3 7 9 3
4 1 5 4
5 5 5 5
> df2
A B C
1 4 4 3
3 3 4 9
6 2 9 3
I need to insert NA-rows in both data frames for each row that does exist in only one of the data frames. In the given example:
> df1
A B C
1 1 5 1
2 3 2 1
3 7 9 3
4 1 5 4
5 5 5 5
6 NA NA NA
> df2
A B C
1 4 4 3
2 NA NA NA
3 3 4 9
4 NA NA NA
5 NA NA NA
6 2 9 3
I will have to perform this operation many times with different data frames, so I need an automatized way to do this. I was trying to solve the issue with different if/else loops, but I am sure there must be a much more automatized way.
We can use functions union, %in% or intersect to find the common rownames and assign rows of an NA dataframe with the values of the dataset if it matches the rownames
un1 <- union(rownames(df1), rownames(df2))
d1 <- as.data.frame(matrix(NA, ncol = ncol(df1),
nrow = length(un1), dimnames = list(un1, names(df1))))
d2 <- d1
d1[rownames(d1) %in% rownames(df1),] <- df1
d2[rownames(d2) %in% rownames(df2),] <- df2
d2
# A B C
#1 4 4 3
#2 NA NA NA
#3 3 4 9
#4 NA NA NA
#5 NA NA NA
#6 2 9 3
I have two dataset x and y
> x
a index b
1 1 1 5
2 NA 2 6
3 2 3 NA
4 NA 4 9
> y
index a
1 2 100
2 4 101
>
I would like to fill the missing values of x with the values contained in y.
I have tried to use the merge function but the result is not what I want.
> merge(x,y, by = 'index', all=T)
index a.x b a.y
1 1 1 5 NA
2 2 NA 6 100
3 3 2 7 NA
4 4 NA 9 101
In the real problem there are additional limitations:
1 - y does not fill all the missing values
2 - x and y have in common more variables (so not only a and index)
EDIT : More realistic example
> x
a index b c
1 1 1 5 NA
2 NA 2 6 NA
3 2 3 NA 5
4 NA 4 9 NA
5 NA 5 10 6
> y
index a c
1 2 100 4
2 4 101 NA
>
The solution would be accepted both in python or R
I used your merge idea and did the following using dplyr. I am sure there will be better ways of doing this task.
index <- 1:5
a <- c(1, NA, 2, NA, NA)
b <- c(5,6,NA,9,10)
c <- c(NA,NA,5,NA,6)
ana <- data.frame(index, a,b,c, stringsAsFactors=F)
index <- c(2,4)
a <- c(100, 101)
c <- c(4, NA)
bob <- data.frame(index, a,c, stringsAsFactors=F)
> ana
index a b c
1 1 1 5 NA
2 2 NA 6 NA
3 3 2 NA 5
4 4 NA 9 NA
5 5 NA 10 6
> bob
index a c
1 2 100 4
2 4 101 NA
ana %>%
merge(., bob, by = "index", all = TRUE) %>%
mutate(a.x = ifelse(a.x %in% NA, a.y, a.x)) %>%
mutate(c.x = ifelse(c.x %in% NA, c.y, c.x))
index a.x b c.x a.y c.y
1 1 1 5 NA NA NA
2 2 100 6 4 100 4
3 3 2 NA 5 NA NA
4 4 101 9 NA 101 NA
5 5 NA 10 6 NA NA
I overwrote a.x (ana$$a) using a.y (bob$a) using mutate. I did a similar thing for c.x (ana$c). If you remove a.y and c.y in the end, that will be the outcome you expect, I think.
Try:
xa = x[,c(1,2)]
m1 = merge(y,xa,all=T)
m1 = m1[!duplicated(m1$index),]
m1$b = x$b[match(m1$index, x$index)]
m1$c = x$c[match(m1$index, x$index)]
m1
index a b c
1 1 1 5 NA
2 2 100 6 NA
4 3 2 NA 5
5 4 101 9 NA
7 5 NA 10 6
or, if there many other columns like b and c:
xa = x[,c(1,2)]
m1 = merge(y,xa,all=T)
m1 = m1[!duplicated(m1$index),]
for(nn in names(x)[3:4]) m1[,nn] = x[,nn][match(m1$index, x$index)]
m1
index a b c
1 1 1 5 NA
2 2 100 6 NA
4 3 2 NA 5
5 4 101 9 NA
7 5 NA 10 6
If there are multiple columns to replace, you could try converting from wide to long form as shown in the first two methods and replace in one step
m1 <- merge(x,y, by="index", all=TRUE)
m1L <- reshape(m1, idvar="index", varying=grep("\\.", colnames(m1)), direction="long", sep=".")
row.names(m1L) <- 1:nrow(m1L)
lst1 <- split(m1L, m1L$time)
indx <- is.na(lst1[[1]][,4:5])
lst1[[1]][,4:5][indx] <- lst1[[2]][,4:5][indx]
res <- lst1[[1]][,c(4,1,2,5)]
res
# a index b c
#1 1 1 5 NA
#2 100 2 6 4
#3 2 3 NA 5
#4 101 4 9 NA
#5 NA 5 10 6
Or you could use dplyr with tidyr
library(dplyr)
library(tidyr)
z <- left_join(x, y, by="index") %>%
gather(Var, Val, matches("\\.")) %>%
separate(Var, c("Var1", "Var2"))
indx1 <- which(is.na(z$Val) & z$Var2=="x")
z$Val[indx1] <- z$Val[indx1+nrow(z)/2]
z %>%
spread(Var1, Val) %>%
filter(Var2=="x") %>%
select(-Var2)
# index b a c
#1 1 5 1 NA
#2 2 6 100 4
#3 3 NA 2 5
#4 4 9 101 NA
#5 5 10 NA 6
Or split the columns by matching names before the . and use lapply to replace the NA's.
indx <- grep("\\.", colnames(m1),value=TRUE)
res <- cbind(m1[!names(m1) %in% indx],
sapply(split(indx, gsub("\\..*", "", indx)), function(x) {
x1 <- m1[x]
indx1 <- is.na(x1[,1])
x1[,1][indx1] <- x1[,2][indx1]
x1[,1]} ))
res
# index b a c
#1 1 5 1 NA
#2 2 6 100 4
#3 3 NA 2 5
#4 4 9 101 NA
#5 5 10 NA 6