I tried the 2 ways below and it seems that whatever I do, a dot is added at the end of the name of the column "repeat":
df <- data.frame(col1=1:5,col2=6:10,"repeat"=11:15)
df <- data.frame(col1=1:5,col2=6:10,`repeat`=11:15)
df
Is there a way to do force it?
Thanks!
After inspecting > data.frame, I found the solution (avoid that the names are checked):
df <- data.frame(col1=1:5,col2=6:10, 'repeat' = 11:15, check.names=FALSE)
df
## col1 col2 repeat
## 1 1 6 11
## 2 2 7 12
## 3 3 8 13
## 4 4 9 14
## 5 5 10 15
An alternative is renaming of the wrongly named df:
df <- data.frame(col1=1:5,col2=6:10, "repeat" = 11:15)
names(df) <- c("col1", "col2", "repeat")
df
## col1 col2 repeat
## 1 1 6 11
## 2 2 7 12
## 3 3 8 13
## 4 4 9 14
## 5 5 10 15
Related
I want to subtract one column from another and create a new one using the corresponding suffix in the first column. I have approx 50 columns
I can do it "manually" as follows...
df$new1 <- df$col_a1 - df$col_b1
df$new2 <- df$col_a2 - df$col_b2
What is the easiest way to create a loop that does the job for me?
We can use grep to identify columns which has "a" and "b" in it and subtract them directly.
a_cols <- grep("col_a", names(df))
b_cols <- grep("col_b", names(df))
df[paste0("new", seq_along(a_cols))] <- df[a_cols] - df[b_cols]
df
# col_a1 col_a2 col_b1 col_b2 new1 new2
#1 10 15 1 5 9 10
#2 9 14 2 6 7 8
#3 8 13 3 7 5 6
#4 7 12 4 8 3 4
#5 6 11 5 9 1 2
#6 5 10 6 10 -1 0
data
Tested on this data
df <- data.frame(col_a1 = 10:5, col_a2 = 15:10, col_b1 = 1:6, col_b2 = 5:10)
Suppose I have the next data frame:
df<-data.frame(step1=c(1,2,3,4),step2=c(5,6,7,8),step3=c(9,10,11,12),step4=c(13,14,15,16))
step1 step2 step3 step4
1 1 5 9 13
2 2 6 10 14
3 3 7 11 15
4 4 8 12 16
and what I have to do is something like the following:
df2<-data.frame(col1=c(1,2,3,4,5,6,7,8,9,10,11,12),col2=c(5,6,7,8,9,10,11,12,13,14,15,16))
col1 col2
1 1 5
2 2 6
3 3 7
4 4 8
5 5 9
6 6 10
7 7 11
8 8 12
9 9 13
10 10 14
11 11 15
12 12 16
How can I do that? consider that more steps can be included (example, 20 steps).
Thanks!!
We can design a function to achieve this task. df_final is the final output. Notice that bin is an argument that the users can specify how many columns to transform together.
# A function to conduct data transformation
trans_fun <- function(df, bin = 3){
# Calculate the number of new columns
new_ncol <- (ncol(df) - bin) + 1
# Create a list to store all data frames
df_list <- lapply(1:new_ncol, function(num){
return(df[, num:(num + bin - 1)])
})
# Convert each data frame to a vector
dt_list2 <- lapply(df_list, unlist)
# Convert dt_list2 to data frame
df_final <- as.data.frame(dt_list2)
# Set the column and row names of df_final
colnames(df_final) <- paste0("col", 1:new_ncol)
rownames(df_final) <- 1:nrow(df_final)
return(df_final)
}
# Apply the trans_fun
df_final <- trans_fun(df)
df_final
col1 col2
1 1 5
2 2 6
3 3 7
4 4 8
5 5 9
6 6 10
7 7 11
8 8 12
9 9 13
10 10 14
11 11 15
12 12 16
Here is a method using dplyr and reshape2 - this assumes all of the columns are the same length.
library(dplyr)
library(reshape2)
Drop the last column from the dataframe
df[,1:ncol(df)-1]%>%
melt() %>%
dplyr::select(col1=value) -> col1
Drop the first column from the dataframe
df %>%
dplyr::select(-step1) %>%
melt() %>%
dplyr::select(col2=value) -> col2
Combine the dataframes
bind_cols(col1, col2)
This should do the work:
df2 <- data.frame(col1 = 1:(length(df$step1) + length(df$step2)))
df2$col1 <- c(df$step1, df$step2, df$step3)
df2$col2 <- c(df$step2, df$step3, df$step4)
Things to point:
The important thing to see in the first line of the code, is the need for creating a table with the right amount of rows
Calling a columns that does not exist will create one, with that name
Deleting columns in R should be done like this df2$col <- NULL
Are you not just looking to do:
df2 <- data.frame(col1 = unlist(df[,-nrow(df)]),
col2 = unlist(df[,-1]))
rownames(df2) <- NULL
df2
col1 col2
1 1 5
2 2 6
3 3 7
4 4 8
5 5 9
6 6 10
7 7 11
8 8 12
9 9 13
10 10 14
11 11 15
12 12 16
I have a data frame which looks as such
A B C
1 3 X1=7;X2=8;X3=9
2 4 X1=10;X2=11;X3=12
5 6 X1=13;X2=14
I would like to parse the C column into separate columns as such...
A B X1 X2 X3
1 3 7 8 9
2 4 10 11 12
5 6 13 14 NA
How would one go about doing this in R?
First, here's the sample data in data.frame form
dd<-data.frame(
A = c(1L, 2L, 5L),
B = c(3L, 4L, 6L),
C = c("X1=7;X2=8;X3=9",
"X1=10;X2=11;X3=12", "X1=13;X2=14"),
stringsAsFactors=F
)
Now I define a small helper function to take vectors like c("A=1","B=2") and changed them into named vectors like c(A="1", B="2").
namev<-function(x) {
a<-strsplit(x,"=")
setNames(sapply(a,'[',2), sapply(a,'[',1))
}
and now I perform the transformations
#turn each row into a named vector
vv<-lapply(strsplit(dd$C,";"), namev)
#find list of all column names
nm<-unique(unlist(sapply(vv, names)))
#extract data from all rows for every column
nv<-do.call(rbind, lapply(vv, '[', nm))
#convert everything to numeric (optional)
class(nv)<-"numeric"
#rejoin with original data
cbind(dd[,-3], nv)
and that gives you
A B X1 X2 X3
1 1 3 7 8 9
2 2 4 10 11 12
3 5 6 13 14 NA
My cSplit function makes solving problems like these fun. Here it is in action:
## Load some packages
library(data.table)
library(devtools) ## Just for source_gist, really
library(reshape2)
## Load `cSplit`
source_gist("https://gist.github.com/mrdwab/11380733")
First, split your values up and create a "long" dataset:
ddL <- cSplit(cSplit(dd, "C", ";", "long"), "C", "=")
ddL
# A B C_1 C_2
# 1: 1 3 X1 7
# 2: 1 3 X2 8
# 3: 1 3 X3 9
# 4: 2 4 X1 10
# 5: 2 4 X2 11
# 6: 2 4 X3 12
# 7: 5 6 X1 13
# 8: 5 6 X2 14
Next, use dcast.data.table (or just dcast) to go from "long" to "wide":
dcast.data.table(ddL, A + B ~ C_1, value.var="C_2")
# A B X1 X2 X3
# 1: 1 3 7 8 9
# 2: 2 4 10 11 12
# 3: 5 6 13 14 NA
Here's one possible approach:
dat <- read.table(text="A B C
1 3 X1=7;X2=8;X3=9
2 4 X1=10;X2=11;X3=12
5 6 X1=13;X2=14", header=TRUE, stringsAsFactors = FALSE)
library(qdapTools)
dat_C <- strsplit(dat$C, ";")
dat_C2 <- sapply(dat_C, function(x) {
y <- strsplit(x, "=")
rep(sapply(y, "[", 1), as.numeric(sapply(y, "[", 2)))
})
data.frame(dat[, -3], mtabulate(dat_C2))
## A B X1 X2 X3
## 1 1 3 7 8 9
## 2 2 4 10 11 12
## 3 5 6 13 14 0
EDIT To obtain the NA values
m <- mtabulate(dat_C2)
m[m==0] <- NA
data.frame(dat[, -3], m)
Here's a nice, somewhat hacky way to get you there.
## read your data
> dat <- read.table(h=T, text = "A B C
1 3 X1=7;X2=8;X3=9
2 4 X1=10;X2=11;X3=12
5 6 X1=13;X2=14", stringsAsFactors = FALSE)
## ---
> s <- strsplit(dat$C, ";|=")
> xx <- unique(unlist(s)[grepl('[A-Z]', unlist(s))])
> sap <- t(sapply(seq(s), function(i){
wh <- which(!xx %in% s[[i]]); n <- suppressWarnings(as.numeric(s[[i]]))
nn <- n[!is.na(n)]; if(length(wh)){ append(nn, NA, wh-1) } else { nn }
})) ## see below for explanation
> data.frame(dat[1:2], sap)
# A B X1 X2 X3
# 1 1 3 7 8 9
# 2 2 4 10 11 12
# 3 5 6 13 14 NA
Basically what's happening in sap is
check which values are missing
change each list element of s to numeric
remove the NA values from (2)
insert NA into the correct position with append
transpose the result
I have a data frame, called df, which contains 4000 values. I have a list of 1000 column numbers, in a data frame called list, which is 1000 rows by 1 column. How can I keep the rows with the numbers in list in the data frame df and throw the rest out. I already tried using:
listv <- as.vector(list)
and then using
dfnew <- df[,listv]
but I get the error
Error in .subset(x, j) : invalid subscript type 'list'
You're mixing up rows and columns subsetting. Here is a minimal example:
df <- data.frame(matrix(1:21, ncol = 3))
df
# X1 X2 X3
# 1 1 8 15
# 2 2 9 16
# 3 3 10 17
# 4 4 11 18
# 5 5 12 19
# 6 6 13 20
# 7 7 14 21
list <- data.frame(V1 = c(1, 4, 6))
list
# V1
# 1 1
# 2 4
# 3 6
df[list[, 1], ]
# X1 X2 X3
# 1 1 8 15
# 4 4 11 18
# 6 6 13 20
df[unlist(list), ]
# X1 X2 X3
# 1 1 8 15
# 4 4 11 18
# 6 6 13 20
Note also that as.vector(list) doesn't create a vector, as you thought it would. You need unlist here (as I used in the last example).
This may seem an obvious questions for someone who has practice with reshape package but I'm trying to get use to its functions and I can't figure out the right syntax!
Let's have the following data frame,
df <- data.frame(matrix(1:12,ncol=3),row.names=letters[1:4])
X1 X2 X3
a 1 5 9
b 2 6 10
c 3 7 11
d 4 8 12
how can we bind the rows into columns in order to get the following result?
X1.a X2.a X3.a X1.b X2.b X3.b X1.c X2.c X3.c X1.d X2.d X3.d
1 5 9 2 6 10 3 7 11 4 8 12
Thank you
This too would work:
vec <- c(t(df))
names(vec) <- c(outer(colnames(df), rownames(df), paste, sep="."))
## > vec
## X1.a X2.a X3.a X1.b X2.b X3.b X1.c X2.c X3.c X1.d X2.d X3.d
## 1 5 9 2 6 10 3 7 11 4 8 12
Since you want it as a vector, there's no need for reshape perse. You can just unlist it and then use setNames to set the names accordingly.
df.t <- as.data.frame(t(df))
vec <- unlist(df.t, use.names=FALSE) # gives a vector not matrix/data.frame
vec.names <- do.call(paste, c(expand.grid(rownames(df.t), colnames(df.t)), sep="."))
vec <- setNames(vec, vec.names)
# X1.a X2.a X3.a X1.b X2.b X3.b X1.c X2.c X3.c X1.d X2.d X3.d
# 1 5 9 2 6 10 3 7 11 4 8 12
Here's one:
m <- melt(cbind(df, rn=rownames(df)), id.vars='rn')
cast(m, ~ rn + variable)
## value a_X1 a_X2 a_X3 b_X1 b_X2 b_X3 c_X1 c_X2 c_X3 d_X1 d_X2 d_X3
## 1 (all) 1 5 9 2 6 10 3 7 11 4 8 12
Or as Arun indicates, acast gives a matrix (without the additional value column):
acast(m, . ~ variable+rn)
## X1_a X1_b X1_c X1_d X2_a X2_b X2_c X2_d X3_a X3_b X3_c X3_d
## [1,] 1 2 3 4 5 6 7 8 9 10 11 12
(Note that the permutation is in the other order, due to the formula being flipped.)