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Replace all particular values in a data frame
Using regex to set a specific digit to NA?Remove Characters in Rparsing quotes out of “NA” stringsReplace specific characters in a variable in data frame in RRegression Analysis in RReplace character in the whole dataframeReplace values in an R data frame in Python using rpy2Join values when creating boxplotHow to replace a character by a newline in Vim?How to replace all occurrences of a string in JavaScriptHow to join (merge) data frames (inner, outer, left, right)Convert data.frame columns from factors to charactersR - list to data frameDrop data frame columns by nameRemove rows with all or some NAs (missing values) in data.frameCreate an empty data.frameHow to get a value from a cell of a dataframe?How to append rows to an R data frame
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
Having a data frame, how do I go about replacing all particular values along all rows and columns. Say for example I want to replace all empty records with NA's (without typing the positions):
df <- data.frame(list(A=c("", "xyz", "jkl"), B=c(12, "", 100)))
A B
1 12
2 xyz
3 jkl 100
Expected result:
A B
1 NA 12
2 xyz NA
3 jkl 100
r dataframe replace
edited Dec 5 '17 at 3:12
Ronak Shah
45.5k104267
asked Oct 21 '13 at 19:41
zxzakzxzak
3,91521922
add a comment |
Having a data frame, how do I go about replacing all particular values along all rows and columns. Say for example I want to replace all empty records with NA's (without typing the positions):
df <- data.frame(list(A=c("", "xyz", "jkl"), B=c(12, "", 100)))
A B
1 12
2 xyz
3 jkl 100
Expected result:
A B
1 NA 12
2 xyz NA
3 jkl 100
r dataframe replace
edited Dec 5 '17 at 3:12
Ronak Shah
45.5k104267
asked Oct 21 '13 at 19:41
zxzakzxzak
3,91521922
add a comment |
Having a data frame, how do I go about replacing all particular values along all rows and columns. Say for example I want to replace all empty records with NA's (without typing the positions):
df <- data.frame(list(A=c("", "xyz", "jkl"), B=c(12, "", 100)))
A B
1 12
2 xyz
3 jkl 100
Expected result:
A B
1 NA 12
2 xyz NA
3 jkl 100
r dataframe replace
edited Dec 5 '17 at 3:12
Ronak Shah
45.5k104267
asked Oct 21 '13 at 19:41
zxzakzxzak
3,91521922
Having a data frame, how do I go about replacing all particular values along all rows and columns. Say for example I want to replace all empty records with NA's (without typing the positions):
df <- data.frame(list(A=c("", "xyz", "jkl"), B=c(12, "", 100)))
A B
1 12
2 xyz
3 jkl 100
Expected result:
A B
1 NA 12
2 xyz NA
3 jkl 100
r dataframe replace
r dataframe replace
edited Dec 5 '17 at 3:12
Ronak Shah
45.5k104267
asked Oct 21 '13 at 19:41
zxzakzxzak
3,91521922
edited Dec 5 '17 at 3:12
Ronak Shah
45.5k104267
asked Oct 21 '13 at 19:41
zxzakzxzak
3,91521922
edited Dec 5 '17 at 3:12
Ronak Shah
45.5k104267
edited Dec 5 '17 at 3:12
Ronak Shah
45.5k104267
edited Dec 5 '17 at 3:12
Ronak Shah
45.5k104267
45.5k104267
asked Oct 21 '13 at 19:41
zxzakzxzak
3,91521922
asked Oct 21 '13 at 19:41
zxzakzxzak
3,91521922
asked Oct 21 '13 at 19:41
zxzakzxzak
3,91521922
3,91521922
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
Like this:
> df[df==""]<-NA
> df
A B
1 <NA> 12
2 xyz <NA>
3 jkl 100
answered Oct 21 '13 at 19:44
mripmrip
10.9k22446
11
is there a way to do this efficiently for more than 1 value!?
– PikkuKatja
Mar 11 '15 at 10:23
14
This doesn't work for factors,df[df=="xyz"]<-"abc"will error with "invalid factor level." Is there a more general solution?
– glallen
Sep 2 '15 at 4:22
not working for me. I tried this: dfSmallDiscreteCustomSalary[dfSmallDiscreteCustomSalary$salary=="<=50K"] <- "49K". Still for unique(dfSmallDiscreteCustomSalary$salary) i get: [1] >50K <=50K
– Codious-JR
Nov 5 '15 at 12:24
3
glallen ... if you're trying to modify a factor column with a new value that already a factor, there are probably more clever ways that what I'm about to suggest, but you could df$factorcolumn <- as.character(df$factorcolumn), then make your modification, and finish off by turning it back into a factor again... df$factorcolumn <- as.factor(df$factorcolumn); it'll be complete with your new level and desired value.
– Joshua Eric Turcotte
Oct 25 '17 at 22:54
1
Scala equivalent for this please..
– sriram
Oct 27 '17 at 19:51
|
show 1 more comment
Since PikkuKatja and glallen asked for a more general solution and I cannot comment yet, I'll write an answer. You can combine statements as in:
> df[df=="" | df==12] <- NA
> df
A B
1 <NA> <NA>
2 xyz <NA>
3 jkl 100
For factors, zxzak's code already yields factors:
> df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)))
> str(df)
'data.frame': 3 obs. of 2 variables:
$ A: Factor w/ 3 levels "","jkl","xyz": 1 3 2
$ B: Factor w/ 3 levels "","100","12": 3 1 2
If in trouble, I'd suggest to temporarily drop the factors.
df[] <- lapply(df, as.character)
answered Dec 8 '15 at 1:12
sedotsedot
319312
add a comment |
We can use data.table to get it quickly.
First create df without factors,
df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)), stringsAsFactors=F)
Now you can use
setDT(df)
for (jj in 1:ncol(df)) set(df, i = which(df[[jj]]==""), j = jj, v = NA)
and you can convert it back to a data.frame
setDF(df)
If you only want to use data.frame and keep factors it's more difficult, you need to work with
levels(df$value)[levels(df$value)==""] <- NA
where value is the name of every column. You need to insert it in a loop.
answered Nov 28 '16 at 19:28
skanskan
2,88183461
2
Why would you use an external library for this use case? Why a loop if this can be solved with one line? How does your answer add value beyond the answers already present? I don't intend to be harsh, I think I am missing something, hence the questions.
– sedot
Jun 21 '17 at 23:34
2
It's much faster for large datasets. It adds an alternative so that the user can choose the best for him.
– skan
Jun 22 '17 at 9:12
add a comment |
If you want to replace multiple values in a data frame, looping through all columns might help.
Say you want to replace "" and 100:
na_codes <- c(100, "")
for (i in seq_along(df))
df[[i]][df[[i]] %in% na_codes] <- NA
answered Apr 7 '17 at 2:11
Olivier MaOlivier Ma
683614
add a comment |
Here are a couple dplyr options:
library(dplyr)
# all columns:
df %>%
mutate_all(~na_if(., ''))
# specific column types:
df %>%
mutate_if(is.factor, ~na_if(., ''))
# specific columns:
df %>%
mutate_at(vars(A, B), ~na_if(., ''))
# or:
df %>%
mutate(A = replace(A, A == '', NA))
# replace can be used if you want something other than NA:
df %>%
mutate(A = as.character(A)) %>%
mutate(A = replace(A, A == '', 'used to be empty'))
answered Mar 22 at 1:48
sbhasbha
2,62822226
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
Like this:
> df[df==""]<-NA
> df
A B
1 <NA> 12
2 xyz <NA>
3 jkl 100
answered Oct 21 '13 at 19:44
mripmrip
10.9k22446
11
is there a way to do this efficiently for more than 1 value!?
– PikkuKatja
Mar 11 '15 at 10:23
14
This doesn't work for factors,df[df=="xyz"]<-"abc"will error with "invalid factor level." Is there a more general solution?
– glallen
Sep 2 '15 at 4:22
not working for me. I tried this: dfSmallDiscreteCustomSalary[dfSmallDiscreteCustomSalary$salary=="<=50K"] <- "49K". Still for unique(dfSmallDiscreteCustomSalary$salary) i get: [1] >50K <=50K
– Codious-JR
Nov 5 '15 at 12:24
3
glallen ... if you're trying to modify a factor column with a new value that already a factor, there are probably more clever ways that what I'm about to suggest, but you could df$factorcolumn <- as.character(df$factorcolumn), then make your modification, and finish off by turning it back into a factor again... df$factorcolumn <- as.factor(df$factorcolumn); it'll be complete with your new level and desired value.
– Joshua Eric Turcotte
Oct 25 '17 at 22:54
1
Scala equivalent for this please..
– sriram
Oct 27 '17 at 19:51
|
show 1 more comment
Like this:
> df[df==""]<-NA
> df
A B
1 <NA> 12
2 xyz <NA>
3 jkl 100
answered Oct 21 '13 at 19:44
mripmrip
10.9k22446
11
is there a way to do this efficiently for more than 1 value!?
– PikkuKatja
Mar 11 '15 at 10:23
14
This doesn't work for factors,df[df=="xyz"]<-"abc"will error with "invalid factor level." Is there a more general solution?
– glallen
Sep 2 '15 at 4:22
not working for me. I tried this: dfSmallDiscreteCustomSalary[dfSmallDiscreteCustomSalary$salary=="<=50K"] <- "49K". Still for unique(dfSmallDiscreteCustomSalary$salary) i get: [1] >50K <=50K
– Codious-JR
Nov 5 '15 at 12:24
3
glallen ... if you're trying to modify a factor column with a new value that already a factor, there are probably more clever ways that what I'm about to suggest, but you could df$factorcolumn <- as.character(df$factorcolumn), then make your modification, and finish off by turning it back into a factor again... df$factorcolumn <- as.factor(df$factorcolumn); it'll be complete with your new level and desired value.
– Joshua Eric Turcotte
Oct 25 '17 at 22:54
1
Scala equivalent for this please..
– sriram
Oct 27 '17 at 19:51
|
show 1 more comment
Like this:
> df[df==""]<-NA
> df
A B
1 <NA> 12
2 xyz <NA>
3 jkl 100
answered Oct 21 '13 at 19:44
mripmrip
10.9k22446
Like this:
> df[df==""]<-NA
> df
A B
1 <NA> 12
2 xyz <NA>
3 jkl 100
answered Oct 21 '13 at 19:44
mripmrip
10.9k22446
answered Oct 21 '13 at 19:44
mripmrip
10.9k22446
answered Oct 21 '13 at 19:44
mripmrip
10.9k22446
answered Oct 21 '13 at 19:44
mripmrip
10.9k22446
10.9k22446
11
is there a way to do this efficiently for more than 1 value!?
– PikkuKatja
Mar 11 '15 at 10:23
14
This doesn't work for factors,df[df=="xyz"]<-"abc"will error with "invalid factor level." Is there a more general solution?
– glallen
Sep 2 '15 at 4:22
not working for me. I tried this: dfSmallDiscreteCustomSalary[dfSmallDiscreteCustomSalary$salary=="<=50K"] <- "49K". Still for unique(dfSmallDiscreteCustomSalary$salary) i get: [1] >50K <=50K
– Codious-JR
Nov 5 '15 at 12:24
3
glallen ... if you're trying to modify a factor column with a new value that already a factor, there are probably more clever ways that what I'm about to suggest, but you could df$factorcolumn <- as.character(df$factorcolumn), then make your modification, and finish off by turning it back into a factor again... df$factorcolumn <- as.factor(df$factorcolumn); it'll be complete with your new level and desired value.
– Joshua Eric Turcotte
Oct 25 '17 at 22:54
1
Scala equivalent for this please..
– sriram
Oct 27 '17 at 19:51
|
show 1 more comment
11
is there a way to do this efficiently for more than 1 value!?
– PikkuKatja
Mar 11 '15 at 10:23
14
This doesn't work for factors,df[df=="xyz"]<-"abc"will error with "invalid factor level." Is there a more general solution?
– glallen
Sep 2 '15 at 4:22
not working for me. I tried this: dfSmallDiscreteCustomSalary[dfSmallDiscreteCustomSalary$salary=="<=50K"] <- "49K". Still for unique(dfSmallDiscreteCustomSalary$salary) i get: [1] >50K <=50K
– Codious-JR
Nov 5 '15 at 12:24
3
glallen ... if you're trying to modify a factor column with a new value that already a factor, there are probably more clever ways that what I'm about to suggest, but you could df$factorcolumn <- as.character(df$factorcolumn), then make your modification, and finish off by turning it back into a factor again... df$factorcolumn <- as.factor(df$factorcolumn); it'll be complete with your new level and desired value.
– Joshua Eric Turcotte
Oct 25 '17 at 22:54
1
Scala equivalent for this please..
– sriram
Oct 27 '17 at 19:51
11
11
is there a way to do this efficiently for more than 1 value!?
– PikkuKatja
Mar 11 '15 at 10:23
is there a way to do this efficiently for more than 1 value!?
– PikkuKatja
Mar 11 '15 at 10:23
14
14
This doesn't work for factors,
df[df=="xyz"]<-"abc" will error with "invalid factor level." Is there a more general solution?– glallen
Sep 2 '15 at 4:22
This doesn't work for factors,
df[df=="xyz"]<-"abc" will error with "invalid factor level." Is there a more general solution?– glallen
Sep 2 '15 at 4:22
not working for me. I tried this: dfSmallDiscreteCustomSalary[dfSmallDiscreteCustomSalary$salary=="<=50K"] <- "49K". Still for unique(dfSmallDiscreteCustomSalary$salary) i get: [1] >50K <=50K
– Codious-JR
Nov 5 '15 at 12:24
not working for me. I tried this: dfSmallDiscreteCustomSalary[dfSmallDiscreteCustomSalary$salary=="<=50K"] <- "49K". Still for unique(dfSmallDiscreteCustomSalary$salary) i get: [1] >50K <=50K
– Codious-JR
Nov 5 '15 at 12:24
3
3
glallen ... if you're trying to modify a factor column with a new value that already a factor, there are probably more clever ways that what I'm about to suggest, but you could df$factorcolumn <- as.character(df$factorcolumn), then make your modification, and finish off by turning it back into a factor again... df$factorcolumn <- as.factor(df$factorcolumn); it'll be complete with your new level and desired value.
– Joshua Eric Turcotte
Oct 25 '17 at 22:54
glallen ... if you're trying to modify a factor column with a new value that already a factor, there are probably more clever ways that what I'm about to suggest, but you could df$factorcolumn <- as.character(df$factorcolumn), then make your modification, and finish off by turning it back into a factor again... df$factorcolumn <- as.factor(df$factorcolumn); it'll be complete with your new level and desired value.
– Joshua Eric Turcotte
Oct 25 '17 at 22:54
1
1
Scala equivalent for this please..
– sriram
Oct 27 '17 at 19:51
Scala equivalent for this please..
– sriram
Oct 27 '17 at 19:51
|
show 1 more comment
Since PikkuKatja and glallen asked for a more general solution and I cannot comment yet, I'll write an answer. You can combine statements as in:
> df[df=="" | df==12] <- NA
> df
A B
1 <NA> <NA>
2 xyz <NA>
3 jkl 100
For factors, zxzak's code already yields factors:
> df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)))
> str(df)
'data.frame': 3 obs. of 2 variables:
$ A: Factor w/ 3 levels "","jkl","xyz": 1 3 2
$ B: Factor w/ 3 levels "","100","12": 3 1 2
If in trouble, I'd suggest to temporarily drop the factors.
df[] <- lapply(df, as.character)
answered Dec 8 '15 at 1:12
sedotsedot
319312
add a comment |
Since PikkuKatja and glallen asked for a more general solution and I cannot comment yet, I'll write an answer. You can combine statements as in:
> df[df=="" | df==12] <- NA
> df
A B
1 <NA> <NA>
2 xyz <NA>
3 jkl 100
For factors, zxzak's code already yields factors:
> df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)))
> str(df)
'data.frame': 3 obs. of 2 variables:
$ A: Factor w/ 3 levels "","jkl","xyz": 1 3 2
$ B: Factor w/ 3 levels "","100","12": 3 1 2
If in trouble, I'd suggest to temporarily drop the factors.
df[] <- lapply(df, as.character)
answered Dec 8 '15 at 1:12
sedotsedot
319312
add a comment |
Since PikkuKatja and glallen asked for a more general solution and I cannot comment yet, I'll write an answer. You can combine statements as in:
> df[df=="" | df==12] <- NA
> df
A B
1 <NA> <NA>
2 xyz <NA>
3 jkl 100
For factors, zxzak's code already yields factors:
> df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)))
> str(df)
'data.frame': 3 obs. of 2 variables:
$ A: Factor w/ 3 levels "","jkl","xyz": 1 3 2
$ B: Factor w/ 3 levels "","100","12": 3 1 2
If in trouble, I'd suggest to temporarily drop the factors.
df[] <- lapply(df, as.character)
answered Dec 8 '15 at 1:12
sedotsedot
319312
Since PikkuKatja and glallen asked for a more general solution and I cannot comment yet, I'll write an answer. You can combine statements as in:
> df[df=="" | df==12] <- NA
> df
A B
1 <NA> <NA>
2 xyz <NA>
3 jkl 100
For factors, zxzak's code already yields factors:
> df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)))
> str(df)
'data.frame': 3 obs. of 2 variables:
$ A: Factor w/ 3 levels "","jkl","xyz": 1 3 2
$ B: Factor w/ 3 levels "","100","12": 3 1 2
If in trouble, I'd suggest to temporarily drop the factors.
df[] <- lapply(df, as.character)
answered Dec 8 '15 at 1:12
sedotsedot
319312
answered Dec 8 '15 at 1:12
sedotsedot
319312
answered Dec 8 '15 at 1:12
sedotsedot
319312
answered Dec 8 '15 at 1:12
sedotsedot
319312
319312
add a comment |
add a comment |
We can use data.table to get it quickly.
First create df without factors,
df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)), stringsAsFactors=F)
Now you can use
setDT(df)
for (jj in 1:ncol(df)) set(df, i = which(df[[jj]]==""), j = jj, v = NA)
and you can convert it back to a data.frame
setDF(df)
If you only want to use data.frame and keep factors it's more difficult, you need to work with
levels(df$value)[levels(df$value)==""] <- NA
where value is the name of every column. You need to insert it in a loop.
answered Nov 28 '16 at 19:28
skanskan
2,88183461
2
Why would you use an external library for this use case? Why a loop if this can be solved with one line? How does your answer add value beyond the answers already present? I don't intend to be harsh, I think I am missing something, hence the questions.
– sedot
Jun 21 '17 at 23:34
2
It's much faster for large datasets. It adds an alternative so that the user can choose the best for him.
– skan
Jun 22 '17 at 9:12
add a comment |
We can use data.table to get it quickly.
First create df without factors,
df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)), stringsAsFactors=F)
Now you can use
setDT(df)
for (jj in 1:ncol(df)) set(df, i = which(df[[jj]]==""), j = jj, v = NA)
and you can convert it back to a data.frame
setDF(df)
If you only want to use data.frame and keep factors it's more difficult, you need to work with
levels(df$value)[levels(df$value)==""] <- NA
where value is the name of every column. You need to insert it in a loop.
answered Nov 28 '16 at 19:28
skanskan
2,88183461
2
Why would you use an external library for this use case? Why a loop if this can be solved with one line? How does your answer add value beyond the answers already present? I don't intend to be harsh, I think I am missing something, hence the questions.
– sedot
Jun 21 '17 at 23:34
2
It's much faster for large datasets. It adds an alternative so that the user can choose the best for him.
– skan
Jun 22 '17 at 9:12
add a comment |
We can use data.table to get it quickly.
First create df without factors,
df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)), stringsAsFactors=F)
Now you can use
setDT(df)
for (jj in 1:ncol(df)) set(df, i = which(df[[jj]]==""), j = jj, v = NA)
and you can convert it back to a data.frame
setDF(df)
If you only want to use data.frame and keep factors it's more difficult, you need to work with
levels(df$value)[levels(df$value)==""] <- NA
where value is the name of every column. You need to insert it in a loop.
answered Nov 28 '16 at 19:28
skanskan
2,88183461
We can use data.table to get it quickly.
First create df without factors,
df <- data.frame(list(A=c("","xyz","jkl"), B=c(12,"",100)), stringsAsFactors=F)
Now you can use
setDT(df)
for (jj in 1:ncol(df)) set(df, i = which(df[[jj]]==""), j = jj, v = NA)
and you can convert it back to a data.frame
setDF(df)
If you only want to use data.frame and keep factors it's more difficult, you need to work with
levels(df$value)[levels(df$value)==""] <- NA
where value is the name of every column. You need to insert it in a loop.
answered Nov 28 '16 at 19:28
skanskan
2,88183461
answered Nov 28 '16 at 19:28
skanskan
2,88183461
answered Nov 28 '16 at 19:28
skanskan
2,88183461
answered Nov 28 '16 at 19:28
skanskan
2,88183461
2,88183461
2
Why would you use an external library for this use case? Why a loop if this can be solved with one line? How does your answer add value beyond the answers already present? I don't intend to be harsh, I think I am missing something, hence the questions.
– sedot
Jun 21 '17 at 23:34
2
It's much faster for large datasets. It adds an alternative so that the user can choose the best for him.
– skan
Jun 22 '17 at 9:12
add a comment |
2
Why would you use an external library for this use case? Why a loop if this can be solved with one line? How does your answer add value beyond the answers already present? I don't intend to be harsh, I think I am missing something, hence the questions.
– sedot
Jun 21 '17 at 23:34
2
It's much faster for large datasets. It adds an alternative so that the user can choose the best for him.
– skan
Jun 22 '17 at 9:12
2
2
Why would you use an external library for this use case? Why a loop if this can be solved with one line? How does your answer add value beyond the answers already present? I don't intend to be harsh, I think I am missing something, hence the questions.
– sedot
Jun 21 '17 at 23:34
Why would you use an external library for this use case? Why a loop if this can be solved with one line? How does your answer add value beyond the answers already present? I don't intend to be harsh, I think I am missing something, hence the questions.
– sedot
Jun 21 '17 at 23:34
2
2
It's much faster for large datasets. It adds an alternative so that the user can choose the best for him.
– skan
Jun 22 '17 at 9:12
It's much faster for large datasets. It adds an alternative so that the user can choose the best for him.
– skan
Jun 22 '17 at 9:12
add a comment |
If you want to replace multiple values in a data frame, looping through all columns might help.
Say you want to replace "" and 100:
na_codes <- c(100, "")
for (i in seq_along(df))
df[[i]][df[[i]] %in% na_codes] <- NA
answered Apr 7 '17 at 2:11
Olivier MaOlivier Ma
683614
add a comment |
If you want to replace multiple values in a data frame, looping through all columns might help.
Say you want to replace "" and 100:
na_codes <- c(100, "")
for (i in seq_along(df))
df[[i]][df[[i]] %in% na_codes] <- NA
answered Apr 7 '17 at 2:11
Olivier MaOlivier Ma
683614
add a comment |
If you want to replace multiple values in a data frame, looping through all columns might help.
Say you want to replace "" and 100:
na_codes <- c(100, "")
for (i in seq_along(df))
df[[i]][df[[i]] %in% na_codes] <- NA
answered Apr 7 '17 at 2:11
Olivier MaOlivier Ma
683614
If you want to replace multiple values in a data frame, looping through all columns might help.
Say you want to replace "" and 100:
na_codes <- c(100, "")
for (i in seq_along(df))
df[[i]][df[[i]] %in% na_codes] <- NA
answered Apr 7 '17 at 2:11
Olivier MaOlivier Ma
683614
answered Apr 7 '17 at 2:11
Olivier MaOlivier Ma
683614
answered Apr 7 '17 at 2:11
Olivier MaOlivier Ma
683614
answered Apr 7 '17 at 2:11
Olivier MaOlivier Ma
683614
683614
add a comment |
add a comment |
Here are a couple dplyr options:
library(dplyr)
# all columns:
df %>%
mutate_all(~na_if(., ''))
# specific column types:
df %>%
mutate_if(is.factor, ~na_if(., ''))
# specific columns:
df %>%
mutate_at(vars(A, B), ~na_if(., ''))
# or:
df %>%
mutate(A = replace(A, A == '', NA))
# replace can be used if you want something other than NA:
df %>%
mutate(A = as.character(A)) %>%
mutate(A = replace(A, A == '', 'used to be empty'))
answered Mar 22 at 1:48
sbhasbha
2,62822226
add a comment |
Here are a couple dplyr options:
library(dplyr)
# all columns:
df %>%
mutate_all(~na_if(., ''))
# specific column types:
df %>%
mutate_if(is.factor, ~na_if(., ''))
# specific columns:
df %>%
mutate_at(vars(A, B), ~na_if(., ''))
# or:
df %>%
mutate(A = replace(A, A == '', NA))
# replace can be used if you want something other than NA:
df %>%
mutate(A = as.character(A)) %>%
mutate(A = replace(A, A == '', 'used to be empty'))
answered Mar 22 at 1:48
sbhasbha
2,62822226
add a comment |
Here are a couple dplyr options:
library(dplyr)
# all columns:
df %>%
mutate_all(~na_if(., ''))
# specific column types:
df %>%
mutate_if(is.factor, ~na_if(., ''))
# specific columns:
df %>%
mutate_at(vars(A, B), ~na_if(., ''))
# or:
df %>%
mutate(A = replace(A, A == '', NA))
# replace can be used if you want something other than NA:
df %>%
mutate(A = as.character(A)) %>%
mutate(A = replace(A, A == '', 'used to be empty'))
answered Mar 22 at 1:48
sbhasbha
2,62822226
Here are a couple dplyr options:
library(dplyr)
# all columns:
df %>%
mutate_all(~na_if(., ''))
# specific column types:
df %>%
mutate_if(is.factor, ~na_if(., ''))
# specific columns:
df %>%
mutate_at(vars(A, B), ~na_if(., ''))
# or:
df %>%
mutate(A = replace(A, A == '', NA))
# replace can be used if you want something other than NA:
df %>%
mutate(A = as.character(A)) %>%
mutate(A = replace(A, A == '', 'used to be empty'))
answered Mar 22 at 1:48
sbhasbha
2,62822226
edited Mar 22 at 10:44
answered Mar 22 at 1:48
sbhasbha
2,62822226
answered Mar 22 at 1:48
sbhasbha
2,62822226
answered Mar 22 at 1:48
sbhasbha
2,62822226
2,62822226
add a comment |
add a comment |
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