Take as an example the dataframe below. I need to change the dataframe by keeping only the columns that are in the filter objects.
test <- data.frame(A = c(1,6,1,2,3) , B = c(1,2,1,1,2), C = c(1,7,6,4,1), D = c(1,1,1,1,1))
filter <- c("A", "B", "C", "D")
filter2 <- c("A","B","D")
To do that I'm using this piece of code:
`%ni%` <- Negate(`%in%`)
test <- test[,-which(names(test) %ni% filter2)]
If I use the filter2 object I get what is expected:
A B D
1 1 1 1
2 6 2 1
3 1 1 1
4 2 1 1
5 3 2 1
However, if I use the filter object, I get a dataframe with zero columns:
data frame with 0 columns and 5 rows
I expected to get an untouched dataframe, since filter had all test columns in it. Why does this happen, and how can I write a more reliable code not to get empty dataframes in these situations?
Use ! instead of -
test[,!(names(test) %ni% filter2)]
test[,!(names(test) %ni% filter)]
by wrapping with which and using -, it works only when the length of output of which is greater than 0
> which(names(test) %ni% filter2)
[1] 3
> which(names(test) %ni% filter)
integer(0)
By doing the -, there is no change in the integer(0) case
> -which(names(test) %ni% filter)
integer(0)
> -which(names(test) %ni% filter2)
[1] -3
thus,
> test[integer(0)]
data frame with 0 columns and 5 rows
I think you can simplify the column selection process by subsetting the dataframe with character vector of column names.
test[filter]
# A B C D
#1 1 1 1 1
#2 6 2 7 1
#3 1 1 6 1
#4 2 1 4 1
#5 3 2 1 1
test[filter2]
# A B D
#1 1 1 1
#2 6 2 1
#3 1 1 1
#4 2 1 1
#5 3 2 1
Related
It seems to me that if a column is not included in the conditional part of the ifelse statement, the ifelse statement in a dplyr mutate function does not work as expected:
mdf <- data.frame(a=c(1,2,3), b=c(3,4,5))
# this works:
> mdf %>% mutate(c=ifelse(a==1,0,1))
a b c
1 1 3 0
2 2 4 1
3 3 5 1
# This does not work (expected column c to be equal to a):
> mdf %>% mutate(c=ifelse(0==1,0,a))
a b c
1 1 3 1
2 2 4 1
3 3 5 1
# This does not work either (expected column c to be equal to a):
> mdf %>% mutate(c=ifelse("a" %in% names(.),a,0))
a b c
1 1 3 1
2 2 4 1
3 3 5 1
If using a regular if-statement, it does work:
> mdf %>% mutate(c=if("a" %in% names(.)){a}else{1})
a b c
1 1 3 1
2 2 4 2
3 3 5 3
However, I was hoping to use the ifelse statement, since it has a cleaner syntax. Is there a way to achieve the desired result with the ifelse statement?
I see that the length of the conditional statement determines what is returned. If the condition evaluates to one True/False value, it only returns one value (instead of the entire column). The value returned seems to be the first value of the desired column:
> mdf %>% mutate(c=ifelse("a" %in% names(.),a,0))
a b c
1 1 3 1
2 2 4 1
3 3 5 1
If I increase the lenght of the condition to the number of rows, the ifelse statement will return the entire column:
> mdf %>% mutate(c=ifelse(rep("a", nrow(.)) %in% names(.),a,0))
a b c
1 1 3 1
2 2 4 2
3 3 5 3
I want to subset a data frame by a vector, but replicate the subsetting for each value in the vector:
data = data.frame(A = c(1,2,3,1), B = c(1,2,3,4))
vec = c(1, 1, 1)
subset(data, A %in% vec)
A B
1 1 1
4 1 4
Instead of this result I want this:
A B
1 1 1
4 1 4
1 1 1
4 1 4
1 1 1
4 1 4
If you use the purrr library, you can do
map_df(vec, function(x) subset(data, A == x))
with base R, it would be
do.call("rbind", lapply(vec, function(x) subset(data, A == x)))
You need to expand it, i.e.
df2 <- subset(data, A %in% vec)
df2[rep(rownames(df2), length(vec)),]
# A B
#1 1 1
#4 1 4
#1.1 1 1
#4.1 1 4
#1.2 1 1
#4.2 1 4
One option with data.table:
library(data.table)
setDT(data, key = 'A')[.(vec)]
# A B
#1: 1 1
#2: 1 4
#3: 1 1
#4: 1 4
#5: 1 1
#6: 1 4
Or use merge, which gives cartesian product as you need when there are duplicated values in the merge-by column:
merge(data, data.frame(A = vec))
# A B
#1: 1 1
#2: 1 1
#3: 1 1
#4: 1 4
#5: 1 4
#6: 1 4
Along the lines of a base R split-apply-combine solution would be
do.call(rbind, lapply(vec, function(i) data[data$A == i, ]))
A B
1 1 1
4 1 4
11 1 1
41 1 4
12 1 1
42 1 4
This could be useful if vec contained an uneven mixture of values. This solution could be expensive if there are many repetitions in vec. In that instance, computation can be reduced by combining it with the rep idea in soto's answer as follows.
# count the number of repetitions by unique value
uni <- table(vec)
# extract unique values
temp <- lapply(as.numeric(names(uni)), function(i) data[data$A == i, ])
# combine results, repeating data.frames according to count
do.call(rbind, temp[rep(seq_along(uni), each=uni)])
I have a data frame initial of the following format
> head(initial)
Strings
1 A,A,B,C
2 A,B,C
3 A,A,A,A,A,B
4 A,A,B,C
5 A,B,C
6 A,A,A,A,A,B
and the data frame I want is final
> head(final)
Strings A B C
1 A,A,B,C 2 1 1
2 A,B,C 1 1 1
3 A,A,A,A,A,B 5 1 0
4 A,A,B,C 2 1 1
5 A,B,C 1 1 1
6 A,A,A,A,A,B 5 1 0
to generate the data frames the following codes can be used to keep the number of rows high
initial<-data.frame(Strings=rep(c("A,A,B,C","A,B,C","A,A,A,A,A,B"),100))
final<-data.frame(Strings=rep(c("A,A,B,C","A,B,C","A,A,A,A,A,B"),100),A=rep(c(2,1,5),100),B=rep(c(1,1,1),100),C=rep(c(1,1,0),100))
What is the fastest way I can achieve this? Any help will be greatly appreciated
We can use base R methods for this task. We split the 'Strings' column (strsplit(...)), set the names of the output list with the sequence of rows, stack to convert to data.frame with key/value columns, get the frequency with table, convert to 'data.frame' and cbind with the original dataset.
cbind(df1, as.data.frame.matrix(
table(
stack(
setNames(
strsplit(as.character(df1$Strings),','), 1:nrow(df1))
)[2:1])))
# Strings A B C D
#1 A,B,C,D 1 1 1 1
#2 A,B,B,D,D,D 1 2 0 3
#3 A,A,A,A,B,C,D,D 4 1 1 2
or we can use mtabulate after splitting the column.
library(qdapTools)
cbind(df1, mtabulate(strsplit(as.character(df1$Strings), ',')))
# Strings A B C D
#1 A,B,C,D 1 1 1 1
#2 A,B,B,D,D,D 1 2 0 3
#3 A,A,A,A,B,C,D,D 4 1 1 2
Update
For the new dataset 'initial', the second method works. If we need to use the first method with the correct order, convert to factor class with levels specified as the unique elements of 'ind'.
df1 <- stack(setNames(strsplit(as.character(initial$Strings), ','),
seq_len(nrow(initial))))
df1$ind <- factor(df1$ind, levels=unique(df1$ind))
cbind(initial, as.data.frame.matrix(table(df1[2:1])))
I have a dataframe that looks something like this, where each row represents a samples, and has repeats of the the same strings
> df
V1 V2 V3 V4 V5
1 a a d d b
2 c a b d a
3 d b a a b
4 d d a b c
5 c a d c c
I want to be able to create a new dataframe, where ideally the headers would be the string variables in the previous dataframe (a, b, c, d) and the contents of each row would be the number of occurrences of each the respective variable from
the original dataframe. Using the example from above, this would look like
> df2
a b c d
1 2 1 0 2
2 2 1 1 1
3 2 1 0 1
4 1 1 1 2
5 1 0 3 1
In my actual dataset, there are hundreds of variables, and thousands of samples, so it'd be ideal if I could automatically pull out the names from the original dataframe, and alphabetize them into the headers for the new dataframe.
You may try
library(qdapTools)
mtabulate(as.data.frame(t(df)))
Or
mtabulate(split(as.matrix(df), row(df)))
Or using base R
Un1 <- sort(unique(unlist(df)))
t(apply(df ,1, function(x) table(factor(x, levels=Un1))))
You can stack the columns and then use table:
table(cbind(id = 1:nrow(mydf),
stack(lapply(mydf, as.character)))[c("id", "values")])
# values
# id a b c d
# 1 2 1 0 2
# 2 2 1 1 1
# 3 2 2 0 1
# 4 1 1 1 2
# 5 1 0 3 1
I would like to break a dataset into two frames - one for which the original dataset has duplicate observations based on a condition and one for which the original dataset does not have duplicate observations based on a condition. In the following example, I would like to break the frame into one for which there is only one coder for an observation and one for which there are two coders::
frame <- data.frame(id = c(1,1,1,2,2,3), coder = c("A", "A", "B", "A", "B", "A"), y = c(4,5,4,1,1,2))
frame
For this, I would like to produce, such that:
frame1:
id coder y
1 1 A 4
2 1 A 5
3 1 B 4
4 2 A 1
5 2 B 1
frame2:
6 3 A 2
You can use aggregate to determine the ids you want in each data frame:
cts <- aggregate(coder~id, frame, function(x) length(unique(x)))
cts
# id coder
# 1 1 2
# 2 2 2
# 3 3 1
Then you can subset as appropriate based on this:
subset(frame, id %in% cts$id[cts$coder >= 2])
# id coder y
# 1 1 A 4
# 2 1 A 5
# 3 1 B 4
# 4 2 A 1
# 5 2 B 1
subset(frame, id %in% cts$id[cts$coder < 2])
# id coder y
# 6 3 A 2
You may also try:
indx <- !colSums(!table(frame$coder, frame$id))
frame[frame$id %in% names(indx)[indx],]
# id coder y
#1 1 A 4
#2 1 A 5
#3 1 B 4
#4 2 A 1
#5 2 B 1
frame[frame$id %in% names(indx)[!indx],]
# id coder y
#6 3 A 2
Explanation
table(frame$coder, frame$id)
# 1 2 3
# A 2 1 1
# B 1 1 0 #Here for id 3, B==0
If we Negate that, the result would be a logical index
!table(frame$coder, frame$id).
Do the colSums of the above, which results
# 1 2 3
# 0 0 1
Negate again and get the index for ids and subset those ids which are TRUE
From this you can subset by matching with the names of the ids