I am looking for a way to copy a column "col1" x times and appending each of these copies with one of x strings from a character vector. Example:
df <- data.frame(col1 = c(1,2,3,4,5))
suffix <- c("a", "b", "c")
resulting in:
df_suffix <- data.frame(col1 = c(1,2,3,4,5), col1_a = c(1,2,3,4,5), col1_b = c(1,2,3,4,5), col1_c = c(1,2,3,4,5))
col1 col1_a col1_b col1_c
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
4 4 4 4 4
5 5 5 5 5
You can use paste() to create the new columns inside df, and assign them the values of the first column:
df[,paste(names(df), suffix, sep = "_")] <- df[,1]
# col1 col1_a col1_b col1_c
#1 1 1 1 1
#2 2 2 2 2
#3 3 3 3 3
#4 4 4 4 4
#5 5 5 5 5
Related
I would like to combine two dataframes using crossing, but some have the same columnnames. For that, I would like to add "_nameofdataframe" to these columns. Here are some reproducible dataframes (dput below):
> df1
person V1 V2 V3
1 A 1 3 3
2 B 4 4 5
3 C 2 1 1
> df2
V2 V3
1 2 5
2 1 6
3 1 2
When I run the following code it will return duplicated column names:
library(tidyr)
crossing(df1, df2, .name_repair = "minimal")
#> # A tibble: 9 × 6
#> person V1 V2 V3 V2 V3
#> <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 A 1 3 3 1 2
#> 2 A 1 3 3 1 6
#> 3 A 1 3 3 2 5
#> 4 B 4 4 5 1 2
#> 5 B 4 4 5 1 6
#> 6 B 4 4 5 2 5
#> 7 C 2 1 1 1 2
#> 8 C 2 1 1 1 6
#> 9 C 2 1 1 2 5
As you can see it returns the column names while being duplicated. My desired output should look like this:
person V1 V2_df1 V3_df1 V2_df2 V3_df2
1 A 1 3 3 1 2
2 A 1 3 3 1 6
3 A 1 3 3 2 5
4 B 4 4 5 1 2
5 B 4 4 5 1 6
6 B 4 4 5 2 5
7 C 2 1 1 1 2
8 C 2 1 1 1 6
9 C 2 1 1 2 5
So I was wondering if anyone knows a more automatic way to give the duplicated column names a name like in the desired output above with crossing?
dput of df1 and df2:
df1 <- structure(list(person = c("A", "B", "C"), V1 = c(1, 4, 2), V2 = c(3,
4, 1), V3 = c(3, 5, 1)), class = "data.frame", row.names = c(NA,
-3L))
df2 <- structure(list(V2 = c(2, 1, 1), V3 = c(5, 6, 2)), class = "data.frame", row.names = c(NA,
-3L))
As you probably know, the .name_repair parameter can take a function. The problem is crossing() only passes that function one argument, a vector of the concatenated column names() of both data frames. So we can't easily pass the names of the data frame objects to it. It seems to me that there are two solutions:
Manually add the desired suffix to an anonymous function.
Create a wrapper function around crossing().
1. Manually add the desired suffix to an anonymous function
We can simply supply the suffix as a character vector to the anonymous .name_repair parameter, e.g. suffix = c("_df1", "_df2").
crossing(
df1,
df2,
.name_repair = \(x, suffix = c("_df1", "_df2")) {
names_to_repair <- names(which(table(x) == 2))
x[x %in% names_to_repair] <- paste0(
x[x %in% names_to_repair],
rep(
suffix,
each = length(unique(names_to_repair))
)
)
x
}
)
# person V1 V2_df1 V3_df1 V2_df2 V3_df2
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 A 1 3 3 1 2
# 2 A 1 3 3 1 6
# 3 A 1 3 3 2 5
# 4 B 4 4 5 1 2
# 5 B 4 4 5 1 6
# 6 B 4 4 5 2 5
# 7 C 2 1 1 1 2
# 8 C 2 1 1 1 6
# 9 C 2 1 1 2 5
The disadvantage of this is that there is a room for error when typing the suffix, or that we might forget to change it if we change the names of the data frames.
Also note that we are checking for names which appear twice. If one of your original data frames already has broken (duplicated) names then this function will also rename those columns. But I think it would be unwise to try to do any type of join if either data frame did not have unique column names.
2. Create a wrapper function around crossing()
This might be more in the spirit of the tidyverse. Thecrossing() docs to which you linked state crossing() is a wrapper around expand_grid(). The source for expand_grid() show that it is basically a wrapper which uses map() to apply vctrs::vec_rep() to some inputs. So if we want to add another function to the call stack, there are two ways I can think of:
Using deparse(substitute())
crossing_fix_names <- function(df_1, df_2) {
suffixes <- paste0(
"_",
c(deparse(substitute(df_1)), deparse(substitute(df_2)))
)
crossing(
df_1,
df_2,
.name_repair = \(x, suffix = suffixes) {
names_to_repair <- names(which(table(x) == 2))
x[x %in% names_to_repair] <- paste0(
x[x %in% names_to_repair],
rep(
suffix,
each = length(unique(names_to_repair))
)
)
x
}
)
}
# Output the same as above
crossing_fix_names(df1, df2)
The disadvantage of this is that deparse(substitute()) is ugly and can occasionally have surprising behaviour. The advantage is we do not need to remember to manually add the suffixes.
Using match.call()
crossing_fix_names2 <- function(df_1, df_2) {
args <- as.list(match.call())
suffixes <- paste0(
"_",
c(
args$df_1,
args$df_2
)
)
crossing(
df_1,
df_2,
.name_repair = \(x, suffix = suffixes) {
names_to_repair <- names(which(table(x) == 2))
x[x %in% names_to_repair] <- paste0(
x[x %in% names_to_repair],
rep(
suffix,
each = length(unique(names_to_repair))
)
)
x
}
)
}
# Also the same output
crossing_fix_names2(df1, df2)
As we don't have the drawbacks of deparse(substitute()) and we don't have to manually specify the suffix, I think this is the probably the best approach.
test for the condition using dputs :
colnames(df1) %in% colnames(df2)
[1] FALSE FALSE TRUE TRUE
rename
colnames(df2) <- paste0(colnames(df2), '_df2')
then cbind
cbind(df1,df2)
person V1 V2 V3 V2_df2 V3_df2
1 A 1 3 3 2 5
2 B 4 4 5 1 6
3 C 2 1 1 1 2
not so elegant, but usefully discernible later.
is it possible to filter rows of one dataframe based on another dataframe?
I have this 2 dataframe:
df_node <- data.frame( id= c("a","b","c","d","e","f","g","h","i"),
group= c(1,1,1,2,2,2,3,3,3))
df_link <- data.frame(from = c("a","d","f","i","b"),
to = c("d","f","i","b","h"))
I would like to delete the lines with characters that are not present in the second dataframe, like this:
here is a basic way to do that:
df_node <- data.frame( id= c("a","b","c","d","e","f","g","h","i"),
group= c(1,1,1,2,2,2,3,3,3))
df_link <- data.frame(from = c("a","d","f","i","b"),
to = c("d","f","i","b","h"))
library(dplyr)
df_result <- df_node%>%
filter(id%in%c(df_link$from,df_link$to))
df_result
# > df_result
# id group
# 1 a 1
# 2 b 1
# 3 d 2
# 4 f 2
# 5 h 3
# 6 i 3
We could use a semi_join:
library(dplyr)
df_node |>
semi_join(tibble(id = c(df_link$from, df_link$to)))
Output:
id group
1 a 1
2 b 1
3 d 2
4 f 2
5 h 3
6 i 3
Here is a oneliner with base R:
df_node[df_node$id %in% unlist(df_link),]
id group
1 a 1
2 b 1
4 d 2
6 f 2
8 h 3
9 i 3
But you could also use a join:
library(dplyr)
df_uniqueID <- data.frame(id = unique(c(df_link$from,df_link$to)) )
right_join(df_node,df_uniqueID)
Joining, by = "id"
id group
1 a 1
2 b 1
3 d 2
4 f 2
5 h 3
6 i 3
This question already has answers here:
Dictionary style replace multiple items
(11 answers)
Closed 1 year ago.
Another thread solved a similar problem very nicely
But what i would like to do is get rid of some redundancy in my similar problem.
Using their example:
df <- data.frame(name = rep(letters[1:3], each = 3), foo=rep(1:9),var1 = letters[1:3], var2 = rep(3:5, each = 3))
creates:
df
name foo var1 var2
1 a 1 a 3
2 a 2 a 3
3 a 3 a 3
4 b 4 b 4
5 b 5 b 4
6 b 6 b 4
7 c 7 c 5
8 c 8 c 5
9 c 9 c 5
But what do i need to do to replace multiple characters with unique values?
a=1
b=2
c=3
I tried:
df[,c(4,6)] <- lapply(df[,c(4,6)], function(x) replace(x,x %in% "a", 1),
replace(x,x %in% "b", 2),
replace(x,x %in% "c", 3))
and
z<- c("a","b","c")
y<- c(1,2,3)
df[,c(1,3)] <- lapply(df[,c(1,3)], function(x) replace(x,x %in% z, y))
But neither seem to work.
Thanks.
You can use dplyr::recode
df <- data.frame(name = rep(letters[1:3], each = 3), foo=rep(1:9),var1 = letters[1:3], var2 = rep(3:5, each = 3))
library(dplyr, warn.conflicts = FALSE)
df %>%
mutate(across(c(name, var1), ~ recode(., a = 1, b = 2, c = 3)))
#> name foo var1 var2
#> 1 1 1 1 3
#> 2 1 2 2 3
#> 3 1 3 3 3
#> 4 2 4 1 4
#> 5 2 5 2 4
#> 6 2 6 3 4
#> 7 3 7 1 5
#> 8 3 8 2 5
#> 9 3 9 3 5
Created on 2021-10-19 by the reprex package (v2.0.1)
Across will apply the function defined by ~ recode(., a = 1, b = 2, c = 3) to both name and var1.
Using ~ and . is another way to define a function in across. This function is equivalent to the one defined by function(x) recode(x, a = 1, b = 2, c = 3), and you could use that code in across instead of the ~ form and it would give the same result. The only name I know for this is what it's called in ?across, which is "purrr-style lambda function", because the purrr package was the first to use formulas to define functions in this way.
If you want to see the actual function created by the formula, you can look at rlang::as_function(~ recode(., a = 1, b = 2, c = 3)), although it's a little more complex than the one above to support the use of ..1, ..2 and ..3 which are not used here.
Now that R supports the easier way of defining functions below, this purrr-style function is maybe no longer useful, it's just an old habit to write it that way.
df <- data.frame(name = rep(letters[1:3], each = 3), foo=rep(1:9),var1 = letters[1:3], var2 = rep(3:5, each = 3))
library(dplyr, warn.conflicts = FALSE)
df %>%
mutate(across(c(name, var1), \(x) recode(x, a = 1, b = 2, c = 3)))
#> name foo var1 var2
#> 1 1 1 1 3
#> 2 1 2 2 3
#> 3 1 3 3 3
#> 4 2 4 1 4
#> 5 2 5 2 4
#> 6 2 6 3 4
#> 7 3 7 1 5
#> 8 3 8 2 5
#> 9 3 9 3 5
Created on 2021-10-19 by the reprex package (v2.0.1)
A simple for loop would do the trick:
for (i in 1:length(z)) {
df[df==z[i]] <- y[i]
}
df
name foo var1 var2
1 1 1 1 3
2 1 2 2 3
3 1 3 3 3
4 2 4 1 4
5 2 5 2 4
6 2 6 3 4
7 3 7 1 5
8 3 8 2 5
9 3 9 3 5
You could use a lookup vector combined with apply:
z <- c("a","b","c")
y <- c(1,2,3)
lookup <- setNames(y, z)
df[,c(1,3)] <- apply(df[,c(1,3)], 2, function(x) lookup[x])
df
This returns
name foo var1 var2
1 1 1 1 3
2 1 2 2 3
3 1 3 3 3
4 2 4 1 4
5 2 5 2 4
6 2 6 3 4
7 3 7 1 5
8 3 8 2 5
9 3 9 3 5
If you are open to a tidyverse approach:
library(tidyverse)
df_new <- df %>%
mutate(across(c(var1, name), ~case_when(. == 'a' ~ 1,
. == 'b' ~ 2,
. == 'c' ~ 3)))
df_new
name foo var1 var2
1 1 1 1 3
2 1 2 2 3
3 1 3 3 3
4 2 4 1 4
5 2 5 2 4
6 2 6 3 4
7 3 7 1 5
8 3 8 2 5
9 3 9 3 5
Note, this code works only if you change all values of your column. E.g. if there was a „d“ in your var1 column that you don‘t tuen into a number, it would be changed to NA.
# Import data: df => data.frame
df <- data.frame(name = rep(letters[1:3], each = 3), foo=rep(1:9),var1 = letters[1:3], var2 = rep(3:5, each = 3))
# Function performing a mapping replacement:
# replaceMultipleValues => function()
replaceMultipleValues <- function(df, mapFrom, mapTo){
# Extract the values in the data.frame:
# dfVals => named character vector
dfVals <- unlist(df)
# Get all values in the mapping & data
# and assign a name to them: tmp1 => named character vector
tmp1 <- c(
setNames(mapTo, mapFrom),
setNames(dfVals, dfVals)
)
# Extract the unique values:
# valueMap => named character vector
valueMap <- tmp1[!(duplicated(names(tmp1)))]
# Recode the values, coerce vectors to appropriate
# types: res => data.frame
res <- type.convert(
data.frame(
matrix(
valueMap[dfVals],
nrow = nrow(df),
ncol = ncol(df),
dimnames = dimnames(df)
)
)
)
# Explicitly define the returned object: data.frame => env
return(res)
}
# Recode values in data.frame:
# res => data.frame
res <- replaceMultipleValues(
df,
c("a", "b", "c"),
c("1", "2", "3")
)
# Print data.frame to console:
# data.frame => stdout(console)
res
I have this dataframe (df):
df <- data.frame(Data1 = c(1,3),
Data2 = c(3,9),
Data3 = c(7,2),
Data1Status = c(1,4),
Data2Status = c(2,5),
Data3Status = c(3,6),
NumberOfMaxValue = c(3,2))
Data1 Data2 Data3 Data1Status Data2Status Data3Status NumberOfMaxValue
1 3 7 1 2 3 3
3 9 2 4 5 6 2
And i want do get this new column:
Data1 Data2 Data3 Data1Status Data2Status Data3Status NumberOfMaxValue DataMaxStatus
1 3 7 1 2 3 3 3
3 9 2 4 5 6 2 5
I tried something like this:
DataMaxStatus = df[, as.numeric(df$NumberOfMaxValue) + 3], but it didn't work.
EDIT/EXPLANATION:
NumberOfMaxValue is the number of the biggest data (1, 2 or 3)
DataMaxStatus is the status of the greater number between Data1, Data2 e Data3
We can get the corresponding Status value by creating a matrix of row/column index to subset from Status columns.
cols <- grep('Status', names(df))
df$DataMaxStatus <- df[cols][cbind(1:nrow(df), df$NumberOfMaxValue)]
df
# Data1 Data2 Data3 Data1Status Data2Status Data3Status NumberOfMaxValue DataMaxStatus
#1 1 3 7 1 2 3 3 3
#2 3 9 2 4 5 6 2 5
I have 2 dataframes in R (df1, df2).
A C D
1 1 1
2 2 2
df2 as
A B C
1 1 1
2 2 2
How can I merge these 2 dataframes to produce the following output?
A B C D
2 1 2 1
4 2 4 2
Columns are sorted and column values are added. Both DFs have same number of rows. Thank you in advance.
Code to create DF:
df1 <- data.frame("A" = 1:2, "C" = 1:2, "D" = 1:2)
df2 <- data.frame("A" = 1:2, "B" = 1:2, "C" = 1:2)
nm1 = names(df1)
nm2 = names(df2)
nm = intersect(nm1, nm2)
if (length(nm) == 0){ # if no column names in common
cbind(df1, df2)
} else { # if column names in common
cbind(df1[!nm1 %in% nm2], # columns only in df1
df1[nm] + df2[nm], # add columns common to both
df2[!nm2 %in% nm1]) # columns only in df2
}
# D A C B
#1 1 2 2 1
#2 2 4 4 2
You can try:
library(tidyverse)
list(df2, df1) %>%
map(rownames_to_column) %>%
bind_rows %>%
group_by(rowname) %>%
summarise_all(sum, na.rm = TRUE)
# A tibble: 2 x 5
rowname A B C D
<chr> <int> <int> <int> <int>
1 1 2 1 2 1
2 2 4 2 4 2
By using left_join() from dplyr you won't lose the column
library(tidyverse)
dat1 <- tibble(a = 1:10,
b = 1:10,
c = 1:10)
dat2 <- tibble(c = 1:10,
d = 1:10,
e = 1:10)
left_join(dat1, dat2, by = "c")
#> # A tibble: 10 x 5
#> a b c d e
#> <int> <int> <int> <int> <int>
#> 1 1 1 1 1 1
#> 2 2 2 2 2 2
#> 3 3 3 3 3 3
#> 4 4 4 4 4 4
#> 5 5 5 5 5 5
#> 6 6 6 6 6 6
#> 7 7 7 7 7 7
#> 8 8 8 8 8 8
#> 9 9 9 9 9 9
#> 10 10 10 10 10 10
Created on 2019-01-16 by the reprex package (v0.2.1)
allnames <- sort(unique(c(names(df1), names(df2))))
df3 <- data.frame(matrix(0, nrow = nrow(df1), ncol = length(allnames)))
names(df3) <- allnames
df3[,allnames %in% names(df1)] <- df3[,allnames %in% names(df1)] + df1
df3[,allnames %in% names(df2)] <- df3[,allnames %in% names(df2)] + df2
df3
A B C D
1 2 1 2 1
2 4 2 4 2
Here is a fun base R method with Reduce.
Reduce(cbind,
list(Reduce("+", list(df1[intersect(names(df1), names(df2))],
df2[intersect(names(df1), names(df2))])), # sum results
df1[setdiff(names(df1), names(df2))], # in df1, not df2
df2[setdiff(names(df2), names(df1))])) # in df2, not df1
This returns
A C D B
1 2 2 1 1
2 4 4 2 2
This assumes that both df1 and df2 have columns that are not present in the other. If this is not true, you'd have to adjust the list.
Note also that you could replace Reduce with do.call in both places and you'd get the same result.