I have this data:
dat=list(structure(list(Group.1 = structure(3:4, .Label = c("A","B", "C", "D", "E", "F"), class = "factor"), Pr1 = c(65, 75)), row.names = c(NA, -2L), class = "data.frame"),NULL, structure(list( Group.1 = structure(3:4, .Label = c("A","B", "C", "D", "E", "F"), class = "factor"), Pr1 = c(81,4)), row.names = c(NA,-2L), class = "data.frame"))
I want to use combine using bind_rows(dat) but keeping the index number as a varaible
Output Include Type([[1]] and [[3]])
type Group.1 Pr1
1 1 C 65
2 1 D 75
3 3 C 81
4 3 D 4
data.table solution
use rbindlist() from the data.table-package, which had built-in id-support that respects NULL df's.
library(data.table)
rbindlist( dat, idcol = TRUE )
.id Group.1 Pr1
1: 1 C 65
2: 1 D 75
3: 3 C 81
4: 3 D 4
dplyr - partly solution
bind_rows also has ID-support, but it 'skips' empty elements...
bind_rows( dat, .id = "id" )
id Group.1 Pr1
1 1 C 65
2 1 D 75
3 2 C 81
4 2 D 4
Note that the ID of the third element from dat becomes 2, and not 3.
According to the documentation of bind_rows() you can supply the name for .id argument of the function. When you apply bind_rows() to the list of data.frames the names of the list containing your data.frames are assigned to the identifier column. [EDIT] But there is a problem mentioned by #Wimpel:
names(dat)
NULL
However, supplying the names to the list will do the thing:
names(dat) <- 1:length(dat)
names(dat)
[1] "1" "2" "3"
bind_rows(dat, .id = "type")
type Group.1 Pr1
1 1 C 65
2 1 D 75
3 3 C 81
4 3 D 4
Or in one line, if you prefer:
bind_rows(setNames(dat, seq_along(dat)), .id = "type")
Related
Given two dataframes df1 and df2 as follows:
df1:
df1 <- structure(list(A = 1L, B = 2L, C = 3L, D = 4L, G = 5L), class = "data.frame", row.names = c(NA,
-1L))
Out:
A B C D G
1 1 2 3 4 5
df2:
df2 <- structure(list(Col1 = c("A", "B", "C", "D", "X"), Col2 = c("E",
"Q", "R", "Z", "Y")), class = "data.frame", row.names = c(NA,
-5L))
Out:
Col1 Col2
1 A E
2 B Q
3 C R
4 D Z
5 X Y
I need to rename columns of df1 using df2, except column G since it not in df2's Col1.
I use df2$Col2[match(names(df1), df2$Col1)] based on the answer from here, but it returns "E" "Q" "R" "Z" NA, as you see column G become NA. I hope it keep the original name.
The expected result:
E Q R Z G
1 1 2 3 4 5
How could I deal with this issue? Thanks.
By using na.omit(it's little bit messy..)
colnames(df1)[na.omit(match(names(df1), df2$Col1))] <- df2$Col2[na.omit(match(names(df1), df2$Col1))]
df1
E Q R Z G
1 1 2 3 4 5
I have success to reproduce your error with
df2 <- data.frame(
Col1 = c("H","I","K","A","B","C","D"),
Col2 = c("a1","a2","a3","E","Q","R","Z")
)
The problem is location of df2$Col1 and names(df1) in match.
na.omit(match(names(df1), df2$Col1))
gives [1] 4 5 6 7, which index does not exist in df1 that has length 5.
For df1, we should change order of terms in match, na.omit(match(df2$Col1,names(df1))) gives [1] 1 2 3 4
colnames(df1)[na.omit(match(df2$Col1, names(df1)))] <- df2$Col2[na.omit(match(names(df1), df2$Col1))]
This will works.
A solution using the rename_with function from the dplyr package.
library(dplyr)
df3 <- df2 %>%
filter(Col1 %in% names(df1))
df4 <- df1 %>%
rename_with(.cols = df3$Col1, .fn = function(x) df3$Col2[df3$Col1 %in% x])
df4
# E Q R Z G
# 1 1 2 3 4 5
i have a data and i want to see if my variables they all have unique value in specefic row
let's say i want to analyze row D
my data
Name F S T
A 1 2 3
B 2 3 4
C 3 4 5
D 4 5 6
> TRUE (because all the three variables have unique value)
Second example
Name F S T
A 1 2 3
B 2 3 4
C 3 4 5
D 4 5 4
>False (because F and T have the same value in row D )
In base R do
f1 <- function(dat, ind) {
tmp <- unlist(dat[ind, -1])
length(unique(tmp)) == length(tmp)
}
-testing
> f1(df, 4)
[1] TRUE
> f1(df1, 4)
[1] FALSE
data
df <- structure(list(Name = c("A", "B", "C", "D"), F = 1:4, S = 2:5,
T = 3:6), class = "data.frame", row.names = c(NA, -4L))
df1 <- structure(list(Name = c("A", "B", "C", "D"), F = 1:4, S = 2:5,
T = c(3L, 4L, 5L, 4L)), class = "data.frame", row.names = c(NA,
-4L))
You can use dplyr for this:
df %>%
summarize_at(c(2:ncol(.)), n_distinct) %>%
summarize(if_all(.fns = ~ .x == nrow(df)))
This question already has answers here:
Merging a lot of data.frames [duplicate]
(1 answer)
How do I replace NA values with zeros in an R dataframe?
(29 answers)
Closed 2 years ago.
I want to merge the following 3 data frames and fill the missing values with -1. I think I should use the fct merge() but not exactly know how to do it.
> df1
Letter Values1
1 A 1
2 B 2
3 C 3
> df2
Letter Values2
1 A 0
2 C 5
3 D 9
> df3
Letter Values3
1 A -1
2 D 5
3 B -1
desire output would be:
Letter Values1 Values2 Values3
1 A 1 0 -1
2 B 2 -1 -1 # fill missing values with -1
3 C 3 5 -1
4 D -1 9 5
code:
> dput(df1)
structure(list(Letter = structure(1:3, .Label = c("A", "B", "C"
), class = "factor"), Values1 = c(1, 2, 3)), class = "data.frame", row.names = c(NA,
-3L))
> dput(df2)
structure(list(Letter = structure(1:3, .Label = c("A", "C", "D"
), class = "factor"), Values2 = c(0, 5, 9)), class = "data.frame", row.names = c(NA,
-3L))
> dput(df3)
structure(list(Letter = structure(c(1L, 3L, 2L), .Label = c("A",
"B", "D"), class = "factor"), Values3 = c(-1, 5, -1)), class = "data.frame", row.names = c(NA,
-3L))
You can get data frames in a list and use merge with Reduce. Missing values in the new dataframe can be replaced with -1.
new_df <- Reduce(function(x, y) merge(x, y, all = TRUE), list(df1, df2, df3))
new_df[is.na(new_df)] <- -1
new_df
# Letter Values1 Values2 Values3
#1 A 1 0 -1
#2 B 2 -1 -1
#3 C 3 5 -1
#4 D -1 9 5
A tidyverse way with the same logic :
library(dplyr)
library(purrr)
list(df1, df2, df3) %>%
reduce(full_join) %>%
mutate(across(everything(), replace_na, -1))
Here's a dplyr solution
df1 %>%
full_join(df2, by = "Letter") %>%
full_join(df3, by = "Letter") %>%
mutate_if(is.numeric, function(x) replace_na(x, -1))
output:
Letter Values1 Values2 Values3
<chr> <dbl> <dbl> <dbl>
1 A 1 0 -1
2 B 2 -1 -1
3 C 3 5 -1
4 D -1 9 5
Lets assume there are 2 columns of two huge dataframes (different lengths) like:
df1 df2
A 1 C X
A 1 D X
B 4 C X
A 1 F X
B 4 A X
B 4 B X
C 7 B X
Each time there is a match in the 1st columns, X should be replaced with data from column 2 of df1. If the 1st column of df2 contains Elements, which are still not in the first column of df1 (F, D), X should be replaced with 0.
Hence there is a huge dataframe, a loop in a loop would not be useful.
The solution should look like this:
df1 df2
A 1 C 7
A 1 D 0
B 4 C 7
A 1 F 0
B 4 A 1
B 4 B 4
C 7 B 4
Thank You in advance
As there are duplicate rows in 'df1', we can get the unique rows
df3 <- unique(df1)
Then, use match to get the idnex
i1 <- match(df2$Col1, df3$Col1)
and based on the index, assign
df2$Col2 <- df3$Col2[i1]
If there are no matches, it would be NA, which can be changed to 0
df2$Col2[is.na(df2$Col2)] <- 0
df2
# Col1 Col2
#1 C 7
#2 D 0
#3 C 7
#4 F 0
#5 A 1
#6 B 4
#7 B 4
Or this can be done with data.table by joining on the 'Col1' and assigning the 'Col2' (after removing the Col2 from the second data) with the Col2 from 'df3'
library(data.table)
setDT(df2)[, Col2 := NULL][df3, Col2 := Col2, on = .(Col1)]
data
df1 <- structure(list(Col1 = c("A", "A", "B", "A", "B", "B", "C"), Col2 = c(1,
1, 4, 1, 4, 4, 7)), class = "data.frame", row.names = c(NA, -7L
))
df2 <- structure(list(Col1 = c("C", "D", "C", "F", "A", "B", "B"), Col2 = c("X",
"X", "X", "X", "X", "X", "X")), class = "data.frame", row.names = c(NA,
-7L))
I would like to use R to get all pairs from two column with index. It may need some loop to finish this function. For example, turn two columns with the gene name and index:
a 1,
b 1,
c 1,
d 2,
e 2
into a new matrix
a b 1,
b c 1,
a c 1,
d e 2
Can anyone help?
A tidyverse option using combn on a grouped data.frame:
library(tidyverse)
df %>% group_by(index) %>%
summarise(gene = list(as_data_frame(t(combn(gene, 2))))) %>%
unnest(.sep = '_')
## # A tibble: 4 × 3
## index gene_V1 gene_V2
## <int> <chr> <chr>
## 1 1 a b
## 2 1 a c
## 3 1 b c
## 4 2 d e
The same logic can be replicated in base R:
df2 <- aggregate(gene ~ index, df, function(x){t(combn(x, 2))})
do.call(rbind, apply(df2, 1, data.frame))
## index gene.1 gene.2
## 1 1 a b
## 2 1 a c
## 3 1 b c
## 4 2 d e
Data
df <- structure(list(gene = c("a", "b", "c", "d", "e"), index = c(1L,
1L, 1L, 2L, 2L)), .Names = c("gene", "index"), row.names = c(NA,
-5L), class = "data.frame")
Here is an option using data.table. Convert the 'data.frame' to 'data.table' (setDT(df)), grouped by 'index', we get the combn of 'gene', transpose it and set the names of the 2nd and 3rd column (if needed).
library(data.table)
setnames(setDT(df)[, transpose(combn(gene, 2, FUN = list)),
by = index], 2:3, paste0("gene", 1:2))[]
# index gene1 gene2
#1: 1 a b
#2: 1 a c
#3: 1 b c
#4: 2 d e