How to find and replace min value in dataframe with text in r
i have dataframe with 20 columns and I like to identify the minimum value in each of the column and replace them with text such as "min". Appreciate any help
sample data :
a b c
-0.05 0.31 0.62
0.78 0.25 -0.01
0.68 0.33 -0.04
-0.01 0.30 0.56
0.55 0.28 -0.03
Desired output
a b c
min 0.31 0.62
0.78 min -0.01
0.68 0.33 min
-0.01 0.30 0.56
0.55 0.28 -0.03
r
edited Nov 14 '18 at 20:29
markus
11.1k1031
asked Nov 14 '18 at 20:21
gurtejgurtej
83
add a comment |
i have dataframe with 20 columns and I like to identify the minimum value in each of the column and replace them with text such as "min". Appreciate any help
sample data :
a b c
-0.05 0.31 0.62
0.78 0.25 -0.01
0.68 0.33 -0.04
-0.01 0.30 0.56
0.55 0.28 -0.03
Desired output
a b c
min 0.31 0.62
0.78 min -0.01
0.68 0.33 min
-0.01 0.30 0.56
0.55 0.28 -0.03
r
edited Nov 14 '18 at 20:29
markus
11.1k1031
asked Nov 14 '18 at 20:21
gurtejgurtej
83
add a comment |
i have dataframe with 20 columns and I like to identify the minimum value in each of the column and replace them with text such as "min". Appreciate any help
sample data :
a b c
-0.05 0.31 0.62
0.78 0.25 -0.01
0.68 0.33 -0.04
-0.01 0.30 0.56
0.55 0.28 -0.03
Desired output
a b c
min 0.31 0.62
0.78 min -0.01
0.68 0.33 min
-0.01 0.30 0.56
0.55 0.28 -0.03
r
edited Nov 14 '18 at 20:29
markus
11.1k1031
asked Nov 14 '18 at 20:21
gurtejgurtej
83
i have dataframe with 20 columns and I like to identify the minimum value in each of the column and replace them with text such as "min". Appreciate any help
sample data :
a b c
-0.05 0.31 0.62
0.78 0.25 -0.01
0.68 0.33 -0.04
-0.01 0.30 0.56
0.55 0.28 -0.03
Desired output
a b c
min 0.31 0.62
0.78 min -0.01
0.68 0.33 min
-0.01 0.30 0.56
0.55 0.28 -0.03
r
r
edited Nov 14 '18 at 20:29
markus
11.1k1031
asked Nov 14 '18 at 20:21
gurtejgurtej
83
edited Nov 14 '18 at 20:29
markus
11.1k1031
asked Nov 14 '18 at 20:21
gurtejgurtej
83
edited Nov 14 '18 at 20:29
markus
11.1k1031
edited Nov 14 '18 at 20:29
markus
11.1k1031
edited Nov 14 '18 at 20:29
markus
11.1k1031
11.1k1031
asked Nov 14 '18 at 20:21
gurtejgurtej
83
asked Nov 14 '18 at 20:21
gurtejgurtej
83
asked Nov 14 '18 at 20:21
gurtejgurtej
83
83
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
apply(df, MARGIN=2, FUN=(function(x){x[which.min(x)] <- 'min'; return(x)})
answered Nov 14 '18 at 20:30
emsinkoemsinko
17115
add a comment |
You can apply a function to each column that replaces the minimum value with a string. This returns a matrix which could be converted into a data frame if desired. As IceCreamToucan pointed out, all rows will be of type character since each variable must have the same type:
apply(df, 2, function(x) {
x[x == min(x)] <- 'min'
return(x)
})
a b c
[1,] "min" "0.31" "0.62"
[2,] "0.78" "min" "-0.01"
[3,] "0.68" "0.33" "min"
[4,] "-0.01" "0.3" "0.56"
[5,] "0.55" "0.28" "-0.03"
answered Nov 14 '18 at 20:28
divibisandivibisan
4,49381531
thank you very much
– gurtej
Nov 14 '18 at 20:30
1
It might be better to dodf <- lapply(df, function(x) replace(x, x == min(x), "min"))to get a data frame in return.
– markus
Nov 14 '18 at 20:32
add a comment |
You can use the method below, but know that this converts all your columns to character, since vectors must have elements which all have the same type.
library(dplyr)
df %>%
mutate_all(~ replace(.x, which.min(.x), 'min'))
# a b c
# 1 min 0.31 0.62
# 2 0.78 min -0.01
# 3 0.68 0.33 min
# 4 -0.01 0.3 0.56
# 5 0.55 0.28 -0.03
answered Nov 14 '18 at 20:26
IceCreamToucanIceCreamToucan
9,2611716
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
apply(df, MARGIN=2, FUN=(function(x){x[which.min(x)] <- 'min'; return(x)})
answered Nov 14 '18 at 20:30
emsinkoemsinko
17115
add a comment |
apply(df, MARGIN=2, FUN=(function(x){x[which.min(x)] <- 'min'; return(x)})
answered Nov 14 '18 at 20:30
emsinkoemsinko
17115
add a comment |
apply(df, MARGIN=2, FUN=(function(x){x[which.min(x)] <- 'min'; return(x)})
answered Nov 14 '18 at 20:30
emsinkoemsinko
17115
apply(df, MARGIN=2, FUN=(function(x){x[which.min(x)] <- 'min'; return(x)})
answered Nov 14 '18 at 20:30
emsinkoemsinko
17115
answered Nov 14 '18 at 20:30
emsinkoemsinko
17115
answered Nov 14 '18 at 20:30
emsinkoemsinko
17115
answered Nov 14 '18 at 20:30
emsinkoemsinko
17115
17115
add a comment |
add a comment |
You can apply a function to each column that replaces the minimum value with a string. This returns a matrix which could be converted into a data frame if desired. As IceCreamToucan pointed out, all rows will be of type character since each variable must have the same type:
apply(df, 2, function(x) {
x[x == min(x)] <- 'min'
return(x)
})
a b c
[1,] "min" "0.31" "0.62"
[2,] "0.78" "min" "-0.01"
[3,] "0.68" "0.33" "min"
[4,] "-0.01" "0.3" "0.56"
[5,] "0.55" "0.28" "-0.03"
answered Nov 14 '18 at 20:28
divibisandivibisan
4,49381531
thank you very much
– gurtej
Nov 14 '18 at 20:30
1
It might be better to dodf <- lapply(df, function(x) replace(x, x == min(x), "min"))to get a data frame in return.
– markus
Nov 14 '18 at 20:32
add a comment |
You can apply a function to each column that replaces the minimum value with a string. This returns a matrix which could be converted into a data frame if desired. As IceCreamToucan pointed out, all rows will be of type character since each variable must have the same type:
apply(df, 2, function(x) {
x[x == min(x)] <- 'min'
return(x)
})
a b c
[1,] "min" "0.31" "0.62"
[2,] "0.78" "min" "-0.01"
[3,] "0.68" "0.33" "min"
[4,] "-0.01" "0.3" "0.56"
[5,] "0.55" "0.28" "-0.03"
answered Nov 14 '18 at 20:28
divibisandivibisan
4,49381531
thank you very much
– gurtej
Nov 14 '18 at 20:30
1
It might be better to dodf <- lapply(df, function(x) replace(x, x == min(x), "min"))to get a data frame in return.
– markus
Nov 14 '18 at 20:32
add a comment |
You can apply a function to each column that replaces the minimum value with a string. This returns a matrix which could be converted into a data frame if desired. As IceCreamToucan pointed out, all rows will be of type character since each variable must have the same type:
apply(df, 2, function(x) {
x[x == min(x)] <- 'min'
return(x)
})
a b c
[1,] "min" "0.31" "0.62"
[2,] "0.78" "min" "-0.01"
[3,] "0.68" "0.33" "min"
[4,] "-0.01" "0.3" "0.56"
[5,] "0.55" "0.28" "-0.03"
answered Nov 14 '18 at 20:28
divibisandivibisan
4,49381531
You can apply a function to each column that replaces the minimum value with a string. This returns a matrix which could be converted into a data frame if desired. As IceCreamToucan pointed out, all rows will be of type character since each variable must have the same type:
apply(df, 2, function(x) {
x[x == min(x)] <- 'min'
return(x)
})
a b c
[1,] "min" "0.31" "0.62"
[2,] "0.78" "min" "-0.01"
[3,] "0.68" "0.33" "min"
[4,] "-0.01" "0.3" "0.56"
[5,] "0.55" "0.28" "-0.03"
answered Nov 14 '18 at 20:28
divibisandivibisan
4,49381531
edited Nov 14 '18 at 20:43
answered Nov 14 '18 at 20:28
divibisandivibisan
4,49381531
answered Nov 14 '18 at 20:28
divibisandivibisan
4,49381531
answered Nov 14 '18 at 20:28
divibisandivibisan
4,49381531
4,49381531
thank you very much
– gurtej
Nov 14 '18 at 20:30
1
It might be better to dodf <- lapply(df, function(x) replace(x, x == min(x), "min"))to get a data frame in return.
– markus
Nov 14 '18 at 20:32
add a comment |
thank you very much
– gurtej
Nov 14 '18 at 20:30
1
It might be better to dodf <- lapply(df, function(x) replace(x, x == min(x), "min"))to get a data frame in return.
– markus
Nov 14 '18 at 20:32
thank you very much
– gurtej
Nov 14 '18 at 20:30
thank you very much
– gurtej
Nov 14 '18 at 20:30
1
1
It might be better to do
df <- lapply(df, function(x) replace(x, x == min(x), "min")) to get a data frame in return.– markus
Nov 14 '18 at 20:32
It might be better to do
df <- lapply(df, function(x) replace(x, x == min(x), "min")) to get a data frame in return.– markus
Nov 14 '18 at 20:32
add a comment |
You can use the method below, but know that this converts all your columns to character, since vectors must have elements which all have the same type.
library(dplyr)
df %>%
mutate_all(~ replace(.x, which.min(.x), 'min'))
# a b c
# 1 min 0.31 0.62
# 2 0.78 min -0.01
# 3 0.68 0.33 min
# 4 -0.01 0.3 0.56
# 5 0.55 0.28 -0.03
answered Nov 14 '18 at 20:26
IceCreamToucanIceCreamToucan
9,2611716
add a comment |
You can use the method below, but know that this converts all your columns to character, since vectors must have elements which all have the same type.
library(dplyr)
df %>%
mutate_all(~ replace(.x, which.min(.x), 'min'))
# a b c
# 1 min 0.31 0.62
# 2 0.78 min -0.01
# 3 0.68 0.33 min
# 4 -0.01 0.3 0.56
# 5 0.55 0.28 -0.03
answered Nov 14 '18 at 20:26
IceCreamToucanIceCreamToucan
9,2611716
add a comment |
You can use the method below, but know that this converts all your columns to character, since vectors must have elements which all have the same type.
library(dplyr)
df %>%
mutate_all(~ replace(.x, which.min(.x), 'min'))
# a b c
# 1 min 0.31 0.62
# 2 0.78 min -0.01
# 3 0.68 0.33 min
# 4 -0.01 0.3 0.56
# 5 0.55 0.28 -0.03
answered Nov 14 '18 at 20:26
IceCreamToucanIceCreamToucan
9,2611716
You can use the method below, but know that this converts all your columns to character, since vectors must have elements which all have the same type.
library(dplyr)
df %>%
mutate_all(~ replace(.x, which.min(.x), 'min'))
# a b c
# 1 min 0.31 0.62
# 2 0.78 min -0.01
# 3 0.68 0.33 min
# 4 -0.01 0.3 0.56
# 5 0.55 0.28 -0.03
answered Nov 14 '18 at 20:26
IceCreamToucanIceCreamToucan
9,2611716
edited Nov 14 '18 at 21:06
answered Nov 14 '18 at 20:26
IceCreamToucanIceCreamToucan
9,2611716
answered Nov 14 '18 at 20:26
IceCreamToucanIceCreamToucan
9,2611716
answered Nov 14 '18 at 20:26
IceCreamToucanIceCreamToucan
9,2611716
9,2611716
add a comment |
add a comment |
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