I have a large list with thousands of data frames included in it. These data frames have multiple columns each. Thereby, I want to automatically bind in each of these data frames the columns into one column. This means that they are appended below each other as shown below. Thereafter, I would transform the list to a data frame which would have varying column lengths due to the different number of columns within each element in the original list.
From this:
y1 y2
1 4
2 5
3 6
To this:
y1
1
2
3
4
5
6
This should be done for each element in the list, whereby the solution needs to take into account that there are thousands of different data frames, which cannot be mentioned individually (example):
df1 = data.frame(
X1 = c(1, 2, 3),
X1.2 = c(4, 5, 6)
)
df2 = data.frame(
X2 = c(7, 8, 9),
X2.2 = c(1, 4, 6)
)
df3 = data.frame(
X3 = c(3, 4, 1),
X3.2 = c(8, 3, 5),
X3.3 = c(3, 1, 9)
)
listOfDataframe = list(df1, df2, df3)
Final output:
df_final = data.frame(
X1 = c(1, 2, 3, 4, 5, 6),
X2 = c(7, 8, 9, 1, 4, 6),
X3 = c(3, 4, 1, 8, 3, 5, 3, 1, 9)
)
Another problem underlying this question is that there will be a differing number of rows, which I do not know how to account for in the data frame, as the columns need to have the same length.
Thank you in advance for your help, it is highly appreciated.
Structure of list within R:
We can unlist after looping over the list with lapply
lst1 <- lapply(listOfDataframe, \(x)
setNames(data.frame(unlist(x, use.names = FALSE)), names(x)[1]))
-output
lst1
[[1]]
X1
1 1
2 2
3 3
4 4
5 5
6 6
[[2]]
X2
1 7
2 8
3 9
4 1
5 4
6 6
[[3]]
X3
1 3
2 4
3 1
4 8
5 3
6 5
7 3
8 1
9 9
If we need to convert the list to a single data.frame, use cbind.na from qPCR
do.call(qpcR:::cbind.na, lst1)
X1 X2 X3
1 1 7 3
2 2 8 4
3 3 9 1
4 4 1 8
5 5 4 3
6 6 6 5
7 NA NA 3
8 NA NA 1
9 NA NA 9
Here is a tidyverse solution:
library(dplyr)
library(purrr)
listOfDataframe %>%
map(~.x %>% stack(.)) %>%
map(~.x %>% select(-ind))
[[1]]
values
1 1
2 2
3 3
4 4
5 5
6 6
[[2]]
values
1 7
2 8
3 9
4 1
5 4
6 6
[[3]]
values
1 3
2 4
3 1
4 8
5 3
6 5
7 3
8 1
9 9
Related
df is a test dataframe and I need to sort the last three columns in ascending order (without hardcoding the order).
df <- data.frame(X = c(1, 2, 3, 4, 5),
Z = c(1, 2, 3, 4, 5),
Y = c(1, 2, 3, 4, 5),
A = c(1, 2, 3, 4, 5),
C = c(1, 2, 3, 4, 5),
B = c(1, 2, 3, 4, 5))
Desired output:
> df
X Z Y A B C
1 1 1 1 1 1 1
2 2 2 2 2 2 2
3 3 3 3 3 3 3
4 4 4 4 4 4 4
5 5 5 5 5 5 5
I'm aware of the order() function but I can't seem to find the right way to implement it to get the desired output.
Update:
Base R:
cbind(df[1:3],df[4:6][,order(colnames(df[4:6]))])
First answer:
We could use relocate from dplyr:
https://dplyr.tidyverse.org/reference/relocate.html
It is configured to arrange columns:
Here we relocate by the index.
We take last (index = 6) and put it before (position 5, which is C)
library(dplyr)
df %>%
relocate(6, .before = 5)
An alternative:
library(dplyr)
df %>%
select(order(colnames(df))) %>%
relocate(4:6, .before = 1)
X Z Y A B C
1 1 1 1 1 1 1
2 2 2 2 2 2 2
3 3 3 3 3 3 3
4 4 4 4 4 4 4
5 5 5 5 5 5 5
In base R, a selection on the first columns then sort the last 3 names :
df[, c(names(df)[1:(ncol(df)-3)], sort(names(df)[ncol(df)-2:0]))]
We want to reorder the columns based on the column names, so if we use names(df) as the argument to order, we can reorder the data frame as follows.
The complicating factor is that order() returns a vector of numbers, so if we want to reorder only a subset of the column names, we'll need an approach that retains the original sort order for the first three columns.
We accomplish this by creating a vector of the first 3 column names, the sorted remaining column names using a function that returns the values rather than locations in the vector, and then use this with the [ form of the extract operator.
df <- data.frame(X = c(1, 2, 3, 4, 5),
Z = c(1, 2, 3, 4, 5),
Y = c(1, 2, 3, 4, 5),
A = c(1, 2, 3, 4, 5),
C = c(1, 2, 3, 4, 5),
B = c(1, 2, 3, 4, 5))
df[,c(names(df[1:3]),sort(names(df[4:6])))]
...and the output:
> df[,c(names(df[1:3]),sort(names(df[4:6])))]
X Z Y A B C
1 1 1 1 1 1 1
2 2 2 2 2 2 2
3 3 3 3 3 3 3
4 4 4 4 4 4 4
5 5 5 5 5 5 5
to_order <- seq(ncol(df)) > ncol(df) - 3
df[order(to_order*order(names(df)))]
#> X Z Y A B C
#> 1 1 1 1 1 1 1
#> 2 2 2 2 2 2 2
#> 3 3 3 3 3 3 3
#> 4 4 4 4 4 4 4
#> 5 5 5 5 5 5 5
Created on 2021-12-24 by the reprex package (v2.0.1)
My data is like:
a <- data.frame(a1=c(2,2,1,1,2,2,3,3),
a2=c(5,4,2,2,5,5,6,6),
a3=c(3,1,5,5,7,7,8,8))
Then, i sort the data like:
aa <- a %>%
arrange(desc(a3),desc(a2),desc(a1))
The data looks like:
> aa
a1 a2 a3
1 3 6 8
2 3 6 8
3 2 5 7
4 2 5 7
5 1 2 5
6 1 2 5
7 2 5 3
8 2 4 1
Now i need to group the data by a3, a2 and a1. So, in aa, the rows 1 and 2 will be in one group, and row 3 and 4 will be in one group as well. Now I need to give every group an index, which starts from 1. So, the data should look like below:
> aa
a1 a2 a3 Index
1 3 6 8 1
2 3 6 8 1
3 2 5 7 2
4 2 5 7 2
5 1 2 5 3
6 1 2 5 3
7 2 5 3 4
8 2 4 1 5
So in summarizing, I need to arrange the data in the descending order first, then group it, then give every group an index starting from 1. Could anyone help me out here?
We could potentially use group_indices, but that would also have a reordering issue. Instead, an option is to paste (or str_c - from stringr) on the columns of interest and then match with unique values of pasted string
library(dplyr)
library(stringr)
aa %>%
mutate(Index = str_c(a1, a2, a3),
Index = match(Index, unique(Index)))
Or instead of arrangeing separately, use it with across
library(tidyr)
a %>%
arrange(across(a1:a3, desc)) %>%
unite(Index, everything(), remove = FALSE) %>%
mutate(Index = match(Index, unique(Index)))
Or with .GRP in data.table
library(dplyr)
setDT(aa)[, Index := .GRP, .(a1, a2, a3)]
aa
# a1 a2 a3 Index
#1: 3 6 8 1
#2: 3 6 8 1
#3: 2 5 7 2
#4: 2 5 7 2
#5: 1 2 5 3
#6: 1 2 5 3
#7: 2 5 3 4
#8: 2 4 1 5
Base R:
a_ordered <- with(a, a[rev(order(a1, a2, a3)), ])
a_ordered$idx <- with(a_ordered,
cumsum(abs(c(
0,
diff(as.integer(factor(paste0(
a1, a2, a3
))))
))) + 1)
Data:
a <- data.frame(
a1 = c(2, 2, 1, 1, 2, 2, 3, 3),
a2 = c(5, 4, 2, 2, 5, 5, 6, 6),
a3 = c(3, 1, 5, 5, 7, 7, 8, 8)
)
I have a list of dataframes (my.list)
d1 <- data.frame(ref = c(1, 2, 3), y2 = c(4, 5, 6), y3 = c(7, 8, 9), y4 = c(10, 11, 12))
d2 <- data.frame(ref = c(3, 2, 1), y2 = c(6, 5, 4), y3 = c(9, 8, 1))
my.list <- list(d1, d2)
d1
ref y2 y3 y4
1 1 4 7 10
2 2 5 8 11
3 3 6 9 12
Now I want to add some columns with absolute difference values to each of the dataframes in this list. I would use the following for loop to do this for dataframe d1
for (i in names(d1)[2:length(names(d1))]){
d1[[paste(i, 'abs_diff', sep="_")]] <- abs(d1[,i]-d1[,2])
}
d1 then looks like this:
ref y2 y3 y4 y2_abs_diff y3_abs_diff y4_abs_diff
1 1 4 7 10 0 3 6
2 2 5 8 11 0 3 6
3 3 6 9 12 0 3 6
But how can I now do this in one shot for all dataframes of my.list? I know I should be using 'lapply' for this, but I can't get it to work.
Wee can use lapply to loop over the list and create the new columns by assignment
my.list1 <- lapply(my.list, function(x) {
x[paste0(names(x)[2:length(x)], "abs_diff")] <- abs(x[-1] - x[,2])
x
})
my.list1
#[[1]]
# ref y2 y3 y4 y2abs_diff y3abs_diff y4abs_diff
#1 1 4 7 10 0 3 6
#2 2 5 8 11 0 3 6
#3 3 6 9 12 0 3 6
#[[2]]
# ref y2 y3 y2abs_diff y3abs_diff
#1 3 6 9 0 3
#2 2 5 8 0 3
#3 1 4 1 0 3
NOTE: When there is a single column to take the difference, due to recycling it will recycle the values to do the operation in each of the columns. Otherwise, we can either make the dimensions same by replicating the column or loop (as in the OP's post)
I have a list:
lst <- list(a1=dfa1, a2=dfa2, b1=dfb1, b2=dfb2)
dfa1 <- data.frame(x=c(1:5), y=c(2, 5, 7, 9, 10))
dfa2 <- data.frame(x=c(1:6), y=c(3, 8, 1, 2, 4, 13))
dfb1 <- data.frame(x=c(1:4), y=c(7, 9, 3, 2))
dfb2 <- data.frame(x=c(1:7), y=c(9, 3, 5, 1, 7, 9, 11))
Base on the partial element match 'a' and 'b', I want column bind the dataframem and the new list should look like below:
new_list
$a
x y1 y2
1 1 2 3
2 2 5 8
3 3 7 1
4 4 9 2
5 5 10 4
$b
x y1 y2
1 1 7 9
2 2 9 3
3 3 3 5
4 4 2 1
Here is a method with lapply and Reduce. lapply iterates through the letters "a" and "b" and applies Reduce to the list elements whose names contain the current letter. Reduce applies the merge function to the two data.frames, merging by the variable "x" and adding the desired suffixes with the given argument. Thanks to zx8754's suggestion, I added seq_along(grep(let, names(lst))) to allow the final names of the variables to increase by the number of group members.
myList <- lapply(c("a", "b"), function(let)
setNames(Reduce(function(x, y) merge(x, y, by="x"),
lst[grep(let, names(lst))]),
c("x", paste0("y", seq_along(grep(let, names(lst)))))))
[[1]]
x y1 y2
1 1 2 3
2 2 5 8
3 3 7 1
4 4 9 2
5 5 10 4
[[2]]
x y1 y2
1 1 7 9
2 2 9 3
3 3 3 5
4 4 2 1
To add names to the list it is probably easiest to do this afterward,
names(myList) <- c("a", "b")
You could also start with the vector
myVec <- c("a", "b")
and then use it in the lapply and in the names line.
Say I have a longitudinal data set as below
ID <- c(1, 1, 2, 2, 3, 3, 4, 4)
time <- c(1, 2, 1, 2, 1, 2, 1, 2)
value <- c(7, 5, 9, 2, NA, 3, 7, NA)
mydata <- data.frame(ID, time, value)
ID time value
1 1 1 7
2 1 2 5
3 2 1 9
4 2 2 2
5 3 1 NA
6 3 2 3
7 4 1 7
8 4 2 NA
In this data-set, we have 4 cases with data at two time-points (let's say pre and post treatment)
Something I want to do is set criteria to delete any case that are not complete for both time-points. In this example, I would want to delete ID3 (who is missing timepoint 1), and ID4 (who is missing timepoint 2). Like below:
ID time value
1 1 1 7
2 1 2 5
3 2 1 9
4 2 2 2
I am not having much luck. I've tried variants of complete.cases() or which() to no avail
I'm still new to R, and would be hugely appreciative if anyone could help me out
Edit: Thank you Ronak for answering my question. Upon reflection of my real data, I have encountered a second problem. My actual data is more reflected by the below:
ID <- c(1, 1, 2, 2, 3, 3, 4, 4, 5, 6, 7, 8)
time <- c(1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1)
value <- c(7, 5, 9, 2, NA, 3, 7, NA, 8, 9, 7, 6)
mydata <- data.frame(ID, time, value)
ID time value
1 1 1 7
2 1 2 5
3 2 1 9
4 2 2 2
5 3 1 NA
6 3 2 3
7 4 1 7
8 4 2 NA
9 5 1 8
10 6 1 9
11 7 1 7
12 8 1 6
Where I would also want to remove cases 5, 6, 7 and 8. These IDs have an entry for Time 1, but not Time 2. Hopefully this makes sense
Thanks a heap
If you switch your data to wide format (where each time point is represented as its own column), then you can use na.omit. Using dplyr and tidyr functions:
library(dplyr)
mydata <- mydata %>%
tidyr::spread(key=time, value=value) %>% # reformat to wide
na.omit() %>% # delete cases with missingness on any variable (i.e. any time point)
tidyr::gather(key="time", value="value", -ID) # put it back in long format
> mydata
ID time value
1 1 1 7
2 2 1 9
3 1 2 5
4 2 2 2
Note that this will work (it will keep only cases with complete data for both time 1 and time 2) even when you have a time point missing without an explicit NA present in the data, like this:
> mydata
ID time value
1 1 1 7
2 1 2 5
3 2 1 9
4 2 2 2
5 3 1 NA
6 3 2 3
7 4 1 7
8 4 2 NA
9 5 1 8
10 6 1 9
11 7 1 7
12 8 1 6
You can do this easily with sqldf.
library(sqldf)
sqldf(' select * from (select ID, count(*) as cnt from mydata where value is not null group by id having cnt >1 ) t1 inner join mydata t2 on t1.ID=t2.ID')
You would select those id having a count greater than 1 and who doesn't have NA in their values and then join back with the original data.
#Ronak already provided
mydata[!mydata$ID %in% mydata$ID[is.na(mydata$value)], ]
For the second part, you can just group over each id and filter on their frequency
k2 <- data.frame(table(mydata$ID))
k2$Var1[k2$Freq > 1]
and then do something like
mydata[mydata$ID %in% k2$Var1[k2$Freq > 1],]
See the updated answer
# Eliminates ID cases with NA
mydata = mydata[!mydata$ID %in% mydata[!complete.cases(mydata) ,]$ID, ]
library(plyr)
# counts all the IDs
cnt = count(mydata, "ID")
# Eliminates any ID that doesn't have 2 observations
mydata[mydata$ID %in% cnt[cnt$freq == 2, ]$ID, ]
ID time value
1 1 1 7
2 1 2 5
3 2 1 9
4 2 2 2