I have a data frame with similar colnames.
I want to calculate rowMeans of columns A and B.
How can I do rowMeans between all A and B columns?
df <- data.frame(A1=c(1,2),A2=c(3,4),A3=c(5,6),A4=c(7,7),A5=c(8,8),A6=c(9,9))
colnames(df)<- c("A","A","B","B","B","C")
An option would be split by the similar column names into a list and then get the rowMeans
i1 <- grep("^(A|B)", names(df))
sapply(split.default(df[i1], names(df)[i1]), rowMeans)
# A B
#[1,] 2 6.666667
#[2,] 3 7.000000
We can iterate over unique names, subset them from original dataframe and take rowMeans.
sapply(c("A", "B"), function(x) rowMeans(df[,colnames(df) == x]))
# A B
#[1,] 2 6.67
#[2,] 3 7.00
An other option using the tidyverse:
library(tidyverse)
df[, "rn"] <- 1:nrow(df)
df %>%
gather(letter, value, -rn) %>%
mutate(letter = str_extract(letter, "[:alpha:]")) %>%
group_by(letter, rn) %>%
summarize(sum = mean(value)) %>%
filter(letter %in% c("A", "B"))
#> # A tibble: 4 x 3
#> # Groups: letter [2]
#> letter rn sum
#> <chr> <int> <dbl>
#> 1 A 1 2
#> 2 A 2 3
#> 3 B 1 6.67
#> 4 B 2 7
You would simply need to submit the dataframe by the columns you want, and then apply the rowMeans() function.
df <- data.frame(A1=c(1,2),A2=c(3,4),A3=c(5,6),A4=c(7,7),A5=c(8,8),A6=c(9,9))
colnames(df)<- c("A","A","B","B","B","C")
rowSums(df[,which(colnames(df) %in% c("A","B"))])
#[1] 24 27
However, as r2evans pointed out in the comment, you should avoid columns with the same names. You would just want to get the position of the columns that determine the start and end of the number of columns between and subset.
colnames(df) <- c(paste0("A",1:2), paste0("B", 1:3), "C1")
strt <- which(colnames(df) == "A1")
end <- which(colnames(df) == "B3")
columrange <- strt:end
rowSums(df[,columrange])
#[1] 24 27
There are many ways to subset by column names. If you didn't rename your columns in your example, you could use grepl() to find them:
df[,grepl("A",colnames(df)) | grepl("B",colnames(df))]
# A1 A2 B1 B2 B3
#1 1 3 5 7 8
#2 2 4 6 7 8
Related
I want to paste a number and some letters together to index them. The columns of my dataframe are as follows;
When CNTR is NA, i want it to be the booking number + an index, so for booking 202653 for example, I want it to be 202653A and 202653B. I already achieved pasting the booking numbers into the CNTR column when its empty with;
dfUNIT$CNTR <- ifelse(is.na(dfUNIT$CNTR), dfUNIT$BOOKING, dfUNIT$CNTR)
which gives me the following table;
But as I said, I need unique CNTR values. My dataframe contains thousands of rows and changes frequently, is there a way to 'index' them the way I want (A, B, C etc)? Thank you in advance
I'll make up some data,
dat <- data.frame(B=c(202658,202654,202653,202653),C=c("TCLU","KOCU",NA,NA))
dplyr
library(dplyr)
dat %>%
group_by(B) %>%
mutate(C = if_else(is.na(C), paste0(B, LETTERS[row_number()]), C))
# # A tibble: 4 x 2
# # Groups: B [3]
# B C
# <dbl> <chr>
# 1 202658 TCLU
# 2 202654 KOCU
# 3 202653 202653A
# 4 202653 202653B
A fundamental risk in this is if you ever have more than 26 rows for a booking, in which case the letter-suffix will fail. An alternative is to append a number instead (e.g., paste0(B, "_", row_number()) or add some other safeguards.
base R alternatives
do.call(rbind, by(dat, dat[,"B",drop=FALSE],
FUN = function(z) transform(z,
C = ifelse(is.na(C), paste0(B, LETTERS[seq_along(z$C)]), C)
)
))
or
append <- ave(dat$C, dat$B, FUN = function(z) ifelse(is.na(z), LETTERS[seq_along(z)], ""))
append
# [1] "" "" "A" "B"
dat$C <- paste0(ifelse(is.na(dat$C), dat$B, dat$C), append)
dat
# B C
# 1 202658 TCLU
# 2 202654 KOCU
# 3 202653 202653A
# 4 202653 202653B
If you don't insist on using letters to index the transformations, here's arough and ready dplyr solution based on rleid from the data.table package:
library(dplyr)
library(data.table)
df %>%
group_by(grp = rleid(B)) %>%
mutate(CNTR_new = if_else(is.na(CNTR), paste0(B, "_", grp), CNTR))
# A tibble: 7 x 4
# Groups: grp [5]
B CNTR grp CNTR_new
<dbl> <chr> <int> <chr>
1 12 TCU 1 TCU
2 13 NA 2 13_2
3 13 NA 2 13_2
4 15 NA 3 15_3
5 1 PVDU 4 PVDU
6 1 NA 4 1_4
7 5 NA 5 5_5
Data:
df <- data.frame(
B = c(12,13,13,15,1,1,5),
CNTR = c("TCU", NA, NA, NA, "PVDU", NA, NA)
)
I need to get the common columns of a data frame list separated in different data frames. Please look at the following example:
df1 <- data.frame(Dates = c('01-01-2020','02-01-2020','03-01-2020'), col1 = c(1,2,3), col2 = c(3,2,1))
df2 <- data.frame(Dates = c('01-01-2020','02-01-2020','03-01-2020'), col1 = c(4,5,6), col2 = c(6,5,4))
df3 <- data.frame(Dates = c('01-01-2020','02-01-2020'), col1 = c(7,8), col2 = c(8,7))
ldf <- list(df1, df2, df3)
The desired output would be the following two data frames:
df_col1:
Date df1 df2 df3
01-01-2020 1 4 7
02-01-2020 2 5 8
03-01-2020 3 6 NA
df_col2:
Date df1 df2 df3
01-01-2020 3 6 8
02-01-2020 2 5 7
03-01-2020 1 4 NA
Of course, ldf is actually way longer, but the number of columns is fixed to 5, so the number of outputs is also fixed (4). This means I wouldn't mind if I use a block of code for each output.
I've tried several things but none seems to work. I'm using base R and hope to find a solution wihtout additional packages.
Thanks a lot for your time!
We bind the list elements with bind_rows from dplyr, then loop over the 'col' columns, along with the common 'Dates', reshape to 'wide' format with pivot_wider and rename if needed
library(dplyr)
library(purrr)
library(tidyr)
library(stringr)
newdf <- bind_rows(ldf)
out <- map(names(newdf)[-1], ~
newdf %>%
select(Dates, .x) %>%
mutate(rn = rowid(Dates)) %>%
pivot_wider(names_from =rn, values_from = !! rlang::sym(.x)) %>%
rename_at(-1, ~ str_c('df', seq_along(.))))
-output
out
#[[1]]
# A tibble: 3 x 4
# Dates df1 df2 df3
# <chr> <dbl> <dbl> <dbl>
#1 01-01-2020 1 4 7
#2 02-01-2020 2 5 8
#3 03-01-2020 3 6 NA
#[[2]]
# A tibble: 3 x 4
# Dates df1 df2 df3
# <chr> <dbl> <dbl> <dbl>
#1 01-01-2020 3 6 8
#2 02-01-2020 2 5 7
#3 03-01-2020 1 4 NA
Or using base R
newdf <- do.call(rbind, ldf)
f1 <- function(dat, colName) {
lst1 <- split(dat[[colName]], dat$Dates)
do.call(rbind, lapply(lst1, `length<-`, max(lengths(lst1))))
}
f1(newdf, 'col1')
f1(newdf, 'col2')
Another Base R option is to do:
m <- Reduce(function(x,y)merge(x, y, by='Dates', all=TRUE), ldf)
lapply(split.default(m[-1], sub("\\..*", "", names(m[-1]))), cbind, m[1])
Another wordy approach using base R:
#Code
names(ldf) <- paste0('df',1:length(ldf))
#Function
myfun <- function(x) {
y <- reshape(x,direction = 'long',
v.names='col',
idvar = 'Dates',varying = list(2:3))
return(y)
}
z <- do.call(rbind,lapply(ldf,myfun))
z$Data <- gsub("\\..*","",rownames(z))
rownames(z) <- NULL
#Reshape
z2 <- reshape(z,idvar = c('Dates','time'),timevar = 'Data')
#List
List <- split(z2,z2$time)
List
Output:
List
$`1`
Dates time col.df1 col.df2 col.df3
1 01-01-2020 1 1 4 7
2 02-01-2020 1 2 5 8
3 03-01-2020 1 3 6 NA
$`2`
Dates time col.df1 col.df2 col.df3
4 01-01-2020 2 3 6 8
5 02-01-2020 2 2 5 7
6 03-01-2020 2 1 4 NA
I am trying to split a dataframe vertically after certain column. Preferably by name. The first half of the split should remain a dataframe and the second should become a matrix. Here is an example.
pp <- rep(1:4,each=4)
cond <- rep(c("A","B"),each=2)
time <- rep(1:2,8)
value <- rnorm(16,1)
df <- data.frame(pp,cond,time,value)
as.data.frame(df %>%
pivot_wider(names_from = c(time), values_from = value))
pp cond 1 2
1 1 A 0.4121770 2.13178625
2 1 B 2.8638453 -0.64314357
3 2 A 2.2587738 1.74448028
4 2 B 0.2737670 0.89784427
5 3 A 0.5831763 2.37123498
6 3 B 0.5158274 1.40670718
7 4 A -0.6313988 1.06272354
8 4 B 2.0142500 0.01102302
Now I'd like to continue piping and split the cols pp and cond into a new dataframe and cols 1 and 2 into a matrix. Any suggestions?
You can try this :
tidyr::pivot_wider(df, names_from = time, values_from = value) %>%
split.default(rep(c(1, 2), each = 2)) -> data1
#change cols 1 and 2 into matrix.
data1[[2]] <- as.matrix(data1[[2]])
data1
#$`1`
# A tibble: 8 x 2
# pp cond
# <int> <chr>
#1 1 A
#2 1 B
#3 2 A
#4 2 B
#5 3 A
#6 3 B
#7 4 A
#8 4 B
#$`2`
# 1 2
#[1,] 1.4442871 1.43913039
#[2,] 2.0406232 1.48409939
#[3,] 0.7551162 1.91599206
#[4,] 1.8006224 0.06343097
#[5,] -0.4007874 1.16027754
#[6,] 0.7260376 0.01446089
#[7,] 1.0839307 -0.31999653
#[8,] 1.1612264 0.37507161
If you want data as two separate objects instead of a list using the column names. Try :
col1 <- c('pp', 'cond')
col2 <- c('1', '2')
df1 <- tidyr::pivot_wider(df, names_from = time, values_from = value)
data1 <- subset(df1, select = col1)
data2 <- subset(df1, select = col2)
Or
data1 <- df1 %>% dplyr::select(all_of(col1))
data2 <- df1 %>% dplyr::select(all_of(col2)) %>% as.matrix()
Given a dataframe such as,
num <- c(5,10,15,20,25)
letter <- c("A", "B", "A", "C", "B")
thelist <- data.frame(num, letter)
I need to find the indices where the letters are the same.
Output:
A 1 3
B 2 5
C 4
Then, take these indices and find the mean of those indices in num.
Output:
A 10
B 17.5
C 20
I cannot use loops or if statements, I am looking at using a sort of apply, which, etc.
As the objective is to find the mean for each similar 'letter', it is better to group by 'letter' and get the mean of 'num'
library(dplyr)
thelist %>%
group_by(letter) %>%
summarise(num = mean(num))
# A tibble: 3 x 2
# letter num
# <fct> <dbl>
#1 A 10
#2 B 17.5
#3 C 20
or in base R
aggregate(num ~ letter, thelist, mean)
To find the index of the same 'letter', we can split the sequence of rows by 'letter
split(seq_len(nrow(thelist)), thelist$letter)
#$A
#[1] 1 3
#$B
#[1] 2 5
#$C
#[1] 4
Another option using data.table:
library(data.table)
setDT(thelist)[, .(ind = paste(.I, collapse = " "),
mean_num = mean(num)
),
by = letter]
Output:
letter ind mean_num
1: A 1 3 10.0
2: B 2 5 17.5
3: C 4 20.0
I'd use dplyr/tidyverse for this:
# setup
library(tidyverse)
# group by letters then get mean of num
thelist %>%
group_by(letter) %>%
summarise(mean_num = mean(num))
You could also use base R with a for loop:
lets <- unique(thelist$letter)
x <- rep(NA, length(lets))
for(i in 1:3){
x[i] <- mean(thelist$num[thelist$letter %in% lets[i]])
}
x
There is my problem that I can't solve it:
Data:
df <- data.frame(f1=c("a", "a", "b", "b", "c", "c", "c"),
v1=c(10, 11, 4, 5, 0, 1, 2))
data.frame:f1 is factor
f1 v1
a 10
a 11
b 4
b 5
c 0
c 1
c 2
# What I want is:(for example, fetch data with the number of element of some level == 2, then to data.frame)
a b
10 4
11 5
Thanks in advance!
I might be missing something simple here , but the below approach using dplyr works.
library(dplyr)
nlevels = 2
df1 <- df %>%
add_count(f1) %>%
filter(n == nlevels) %>%
select(-n) %>%
mutate(rn = row_number()) %>%
spread(f1, v1) %>%
select(-rn)
This gives
# a b
# <int> <int>
#1 10 NA
#2 11 NA
#3 NA 4
#4 NA 5
Now, if you want to remove NA's we can do
do.call("cbind.data.frame", lapply(df1, function(x) x[!is.na(x)]))
# a b
#1 10 4
#2 11 5
As we have filtered the dataframe which has only nlevels observations, we would have same number of rows for each column in the final dataframe.
split might be useful here to split df$v1 into parts corresponding to df$f1. Since you are always extracting equal length chunks, it can then simply be combined back to a data.frame:
spl <- split(df$v1, df$f1)
data.frame(spl[lengths(spl)==2])
# a b
#1 10 4
#2 11 5
Or do it all in one call by combining this with Filter:
data.frame(Filter(function(x) length(x)==2, split(df$v1, df$f1)))
# a b
#1 10 4
#2 11 5
Here is a solution using unstack :
unstack(
droplevels(df[ave(df$v1, df$f1, FUN = function(x) length(x) == 2)==1,]),
v1 ~ f1)
# a b
# 1 10 4
# 2 11 5
A variant, similar to #thelatemail's solution :
data.frame(Filter(function(x) length(x) == 2, unstack(df,v1 ~ f1)))
My tidyverse solution would be:
library(tidyverse)
df %>%
group_by(f1) %>%
filter(n() == 2) %>%
mutate(i = row_number()) %>%
spread(f1, v1) %>%
select(-i)
# # A tibble: 2 x 2
# a b
# * <dbl> <dbl>
# 1 10 4
# 2 11 5
or mixing approaches :
as_tibble(keep(unstack(df,v1 ~ f1), ~length(.x) == 2))
Using all base functions (but you should use tidyverse)
# Add count of instances
x$len <- ave(x$v1, x$f1, FUN = length)
# Filter, drop the count
x <- x[x$len==2, c('f1','v1')]
# Hacky pivot
result <- data.frame(
lapply(unique(x$f1), FUN = function(y) x$v1[x$f1==y])
)
colnames(result) <- unique(x$f1)
> result
a b
1 10 4
2 11 5
I'd like code this, may it helps for you
library(reshape2)
library(dplyr)
aa = data.frame(v1=c('a','a','b','b','c','c','c'),f1=c(10,11,4,5,0,1,2))
cc = aa %>% group_by(v1) %>% summarise(id = length((v1)))
dd= merge(aa,cc) #get the level
ee = dd[dd$aa==2,] #select number of level equal to 2
ee$id = rep(c(1,2),nrow(ee)/2) # reset index like (1,2,1,2)
dcast(ee, id~v1,value.var = 'f1')
all done!