Let I have the below data frame.
df.open<-c(1,4,5)
df.close<-c(2,8,3)
df<-data.frame(df.open, df.close)
> df
df.open df.close
1 1 2
2 4 8
3 5 3
I wanto change column names which includes "open" with "a" and column names which includes "close" with "b":
Namely I want to obtain the below data frame:
a b
1 1 2
2 4 8
3 5 3
I have a lot of such data frames. The pre values(here it is "df.") are changing but "open" and "close" are fix.
Thanks a lot.
We can create a function for reuse
f1 <- function(dat) {
names(dat)[grep('open$', names(dat))] <- 'a'
names(dat)[grep('close$', names(dat))] <- 'b'
dat
}
and apply on the data
df <- f1(df)
-output
df
a b
1 1 2
2 4 8
3 5 3
if these datasets are in a list
lst1 <- list(df, df)
lst1 <- lapply(lst1, f1)
Thanks to dear #akrun's insightful suggestion as always we can do it in one go. So we create character vectors in pattern and replacement arguments of str_replace to be able to carry out both operations at once. We can assign character vector of either length one or more to each one of them. In case of the latter the length of both vectors should correspond. More to the point as the documentation says:
References of the form \1, \2, etc will be replaced with the contents
of the respective matched group (created by ())
library(dplyr)
library(stringr)
df %>%
rename_with(~ str_replace(., c(".*\\.open", ".*\\.close"), c("a", "b")))
a b
1 1 2
2 4 8
3 5 3
Another base R option using gsub + match + setNames
setNames(
df,
c("a", "b")[match(
gsub("[^open|close]", "", names(df)),
c("open", "close")
)]
)
gives
a b
1 1 2
2 4 8
3 5 3
Related
I have a dataframe data with a lot of columns in the form of
...v1...min ...v1...max ...v2...min ...v2...max
1 a a a a
2 b b b b
3 c c c c
where in place ... there could be any expression.
I would like to create a function createData that takes three arguments:
X: a dataframe,
cols: a vector containing first part of the column, so i.e. c("v1", "v2")
fun: a vector containing second part of the column, so i.e. c("min"), or c("max", "min")
and returns filtered dataframe, so - for example:
createData(X, c("v1"), None) would return this kind of dataframe:
...v1...min ...v1...max
1 a a
2 b b
3 c c
while createData(X, c("v1", "v2"), c("min")) would give me
...v1...min ...v2...min
1 a a
2 b b
3 c c
At this point I decided I need to use i.e. select(contains()) from dplyr package.
createData <- function(data, fun, cols)
{
X %>% select(contains())
return(X)
}
What I struggle with is:
how to filter columns that consist two (or maybe more?) strings, i.e. both var1 and min? I tried going with data[grepl(".*(v1*min|min*v1).*", colnames(data), ignore.case=TRUE)] but it doesn't seem to work and also my expressions aren't fixed - they depend on the vector I pass,
how to filter multiple columns with different names, i.e. c("v1", "v2"), passed in a vector? and how to combine it with the first question?
I don't really need to stick with dplyr package, it was just for the sake of the example. Thanks!
EDIT:
An reproducible example:
data = data.frame(AXv1c2min = c(1,2,3),
subv1trwmax = c(4,5,6),
ss25v2xxmin = c(7,8,9),
cwfv2urttmmax = c(10,11,12))
If you pass a vector to contains, it will function like an OR tag, while multiple select statements will have additive effects. So for your esample data:
We can filter for (v1 OR v2) AND min like this:
library(tidyverse)
data %>%
select(contains(c('v1','v2'))) %>%
select(contains('min'))
AXv1c2min ss25v2xxmin
1 1 7
2 2 8
3 3 9
So as a function where either argument is optional:
createData <- function(data, fun=NULL, cols=NULL) {
if (!is.null(fun)) data <- select(data, contains(fun))
if (!is.null(cols)) data <- select(data, contains(cols))
return(data)
}
A series of examples:
createData(data, cols=c('v1', 'v2'), fun='min')
AXv1c2min ss25v2xxmin
1 1 7
2 2 8
3 3 9
createData(data, cols=c('v1'))
AXv1c2min subv1trwmax
1 1 4
2 2 5
3 3 6
createData(data, fun=c('min'))
AXv1c2min ss25v2xxmin
1 1 7
2 2 8
3 3 9
createData(data, cols=c('v1'), fun=c('min', 'max'))
AXv1c2min subv1trwmax
1 1 4
2 2 5
3 3 6
createData(data, cols=c('v1'), fun=c('max'))
subv1trwmax
1 4
2 5
3 6
It is a very basic question.How can you set the column names of data frame to column index? So if you have 4 columns, column names will be 1 2 3 4. The data frame i am using can have up to 100 columns.
It is not good to name the column names with names that start with numbers. Suppose, we name it as seq_along(D). It becomes unnecessarily complicated when we try to extract a column. For example,
names(D) <- seq_along(D)
D$1
#Error: unexpected numeric constant in "D$1"
In that case, we may need backticks or ""
D$"1"
#[1] 1 2 3
D$`1`
#[1] 1 2 3
However, the [ should work
D[["1"]]
#[1] 1 2 3
I would use
names(D) <- paste0("Col", seq_along(D))
D$Col1
#[1] 1 2 3
Or
D[["Col1"]]
#[1] 1 2 3
data
D <- data.frame(a=c(1,2,3),b=c(4,5,6),c=c(7,8,9),d=c(10,11,12))
Just use names:
D <- data.frame(a=c(1,2,3),b=c(4,5,6),c=c(7,8,9),d=c(10,11,12))
names(D) <- 1:ncol(D) # sequence from 1 through the number of columns
I need to subset a dataframe (df) by a string of columns names that I have created - not sure how to inject this into a subet..?
for example
colstoKeep is a character string:
"col1", "col2", "col3", "col4"
how do I push this into a subset function
df<- df[colstoKeep]
I'm sure this is easy.? because the above doesn't work.
df <- data.frame(A=seq(1:5),B=seq(5:1),C=seq(1:5))
df
colsToKeep <- "\"A\", \"C\""
If I understand your question correctly, your colsToKeep variable is a string as given above. In order to extract the variables, you will have to convert that into a vector. If I've used the right format, you can do that with the following code.
library(magrittr)
colsToKeepVector <-
strsplit(colsToKeep, ",") %>%
unlist() %>%
trimws() %>%
gsub("\"", "", .)
df[colsToKeepVector]
However, if I'm also understanding that you had a vector that you collapsed to a string (paste(..., collapse = ", ")?), I would strongly advise you not to do that.
(Edited to match the string format in the question)
df <- data.frame(A=seq(1:5),B=seq(5:1),C=seq(1:5))
df
A B C
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
cols_to_keep <- c("A","C")
df[,cols_to_keep]
A C
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
This question already has answers here:
Repeat each row of data.frame the number of times specified in a column
(10 answers)
Closed 4 years ago.
Embarrassingly basic question, but if you don't know.. I need to reshape a data.frame of count summarised data into what it would've looked like before being summarised. This is essentially the reverse of {plyr} count() e.g.
> (d = data.frame(value=c(1,1,1,2,3,3), cat=c('A','A','A','A','B','B')))
value cat
1 1 A
2 1 A
3 1 A
4 2 A
5 3 B
6 3 B
> (summry = plyr::count(d))
value cat freq
1 1 A 3
2 2 A 1
3 3 B 2
If you start with summry what is the quickest way back to d? Unless I'm mistaken (very possible), {Reshape2} doesn't do this..
Just use rep:
summry[rep(rownames(summry), summry$freq), c("value", "cat")]
# value cat
# 1 1 A
# 1.1 1 A
# 1.2 1 A
# 2 2 A
# 3 3 B
# 3.1 3 B
A variation of this approach can be found in expandRows from my "SOfun" package. If you had that loaded, you would be able to simply do:
expandRows(summry, "freq")
There is a good table to dataframe function on the R cookbook website that you can modify slightly. The only modifications were changing 'Freq' -> 'freq' (to be consistent with plyr::count) and making sure the rownames were reset as increasing integers.
expand.dft <- function(x, na.strings = "NA", as.is = FALSE, dec = ".") {
# Take each row in the source data frame table and replicate it
# using the Freq value
DF <- sapply(1:nrow(x),
function(i) x[rep(i, each = x$freq[i]), ],
simplify = FALSE)
# Take the above list and rbind it to create a single DF
# Also subset the result to eliminate the Freq column
DF <- subset(do.call("rbind", DF), select = -freq)
# Now apply type.convert to the character coerced factor columns
# to facilitate data type selection for each column
for (i in 1:ncol(DF)) {
DF[[i]] <- type.convert(as.character(DF[[i]]),
na.strings = na.strings,
as.is = as.is, dec = dec)
}
row.names(DF) <- seq(nrow(DF))
DF
}
expand.dft(summry)
value cat
1 1 A
2 1 A
3 1 A
4 2 A
5 3 B
6 3 B
In attempting to answer a question earlier, I ran into a problem that seemed like it should be simple, but I couldn't figure out.
If I have a list of dataframes:
df1 <- data.frame(a=1:3, x=rnorm(3))
df2 <- data.frame(a=1:3, x=rnorm(3))
df3 <- data.frame(a=1:3, x=rnorm(3))
df.list <- list(df1, df2, df3)
That I want to rbind together, I can do the following:
df.all <- ldply(df.list, rbind)
However, I want another column that identifies which data.frame each row came from. I expected to be able to use the deparse(substitute(x)) method (here and elsewhere) to get the name of the relevant data.frame and add a column. This is how I approached it:
fun <- function(x) {
name <- deparse(substitute(x))
x$id <- name
return(x)
}
df.all <- ldply(df.list, fun)
Which returns
a x id
1 1 1.1138062 X[[1L]]
2 2 -0.5742069 X[[1L]]
3 3 0.7546323 X[[1L]]
4 1 1.8358605 X[[2L]]
5 2 0.9107199 X[[2L]]
6 3 0.8313439 X[[2L]]
7 1 0.5827148 X[[3L]]
8 2 -0.9896495 X[[3L]]
9 3 -0.9451503 X[[3L]]
So obviously each element of the list does not contain the name I think it does. Can anyone suggest a way to get what I expected (shown below)?
a x id
1 1 1.1138062 df1
2 2 -0.5742069 df1
3 3 0.7546323 df1
4 1 1.8358605 df2
5 2 0.9107199 df2
6 3 0.8313439 df2
7 1 0.5827148 df3
8 2 -0.9896495 df3
9 3 -0.9451503 df3
Define your list with names and it should give you an .id column with the data.frame name
df.list <- list(df1=df1, df2=df2, df3=df3)
df.all <- ldply(df.list, rbind)
Output:
.id a x
1 df1 1 1.84658809
2 df1 2 -0.01177462
3 df1 3 0.58579469
4 df2 1 -0.64748756
5 df2 2 0.24384614
6 df2 3 0.59012676
7 df3 1 -0.63037679
8 df3 2 -1.17416295
9 df3 3 1.09349618
Then you can know the data.frame name from the column df.all$.id
Edit:
As per #Gary Weissman's comment if you want to generate the names automatically you can do
names(df.list) <- paste0('df',seq_along(df.list)
Using base only, one could try something like:
dd <- lapply(seq_along(df.list), function(x) cbind(df_name = paste0('df',x),df.list[[x]]))
do.call(rbind,dd)
In your definition, df.list does not have names, however, even then the deparse substitute idiom does not appear to work easilty (as lapply calls .Internal(lapply(X, FUN)) -- you would have to look at the source to see if the object name was available and how to get it
Something like
names(df.list) <- paste('df', 1:3, sep = '')
foo <- function(n, .list){
.list[[n]]$id <- n
.list[[n]]
}
a x id
1 1 0.8204213 a
2 2 -0.8881671 a
3 3 1.2880816 a
4 1 -2.2766111 b
5 2 0.3912521 b
6 3 -1.3963381 b
7 1 -1.8057246 c
8 2 0.5862760 c
9 3 0.5605867 c
if you want to use your function, instead of deparse(substitute(x)) use match.call(), and you want the second argument, making sure to convert it to character
name <- as.character(match.call()[[2]])