modify value in each row of a dataframe

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}

I have an input called target and a dataframe:

target <- 3

d <- data.frame(x=c(1,2,3),y=c(4,5,6))



x y  

1 4   

2 5   

3 6

I want for each row:

if the value of column x < target then this value <- 0

if the value of column y > target then this value <- 9

the result:

how could I get this?

I tried to use the function apply but it didn't allow to modify the value in dataframe d.

edited Jan 4 at 13:44

Ronak Shah

48k104269

asked Jan 4 at 13:31

DD chen

217

I think you have a typo in the third row of your expected x output.

– Tim Biegeleisen
Jan 4 at 13:35

add a comment |

I have an input called target and a dataframe:

target <- 3

d <- data.frame(x=c(1,2,3),y=c(4,5,6))



x y  

1 4   

2 5   

3 6

I want for each row:

if the value of column x < target then this value <- 0

if the value of column y > target then this value <- 9

the result:

how could I get this?

I tried to use the function apply but it didn't allow to modify the value in dataframe d.

edited Jan 4 at 13:44

Ronak Shah

48k104269

asked Jan 4 at 13:31

DD chen

217

I think you have a typo in the third row of your expected x output.

– Tim Biegeleisen
Jan 4 at 13:35

add a comment |

I have an input called target and a dataframe:

target <- 3

d <- data.frame(x=c(1,2,3),y=c(4,5,6))



x y  

1 4   

2 5   

3 6

I want for each row:

if the value of column x < target then this value <- 0

if the value of column y > target then this value <- 9

the result:

how could I get this?

I tried to use the function apply but it didn't allow to modify the value in dataframe d.

edited Jan 4 at 13:44

Ronak Shah

48k104269

asked Jan 4 at 13:31

DD chen

217

I have an input called target and a dataframe:

target <- 3

d <- data.frame(x=c(1,2,3),y=c(4,5,6))



x y  

1 4   

2 5   

3 6

I want for each row:

if the value of column x < target then this value <- 0

if the value of column y > target then this value <- 9

the result:

how could I get this?

I tried to use the function apply but it didn't allow to modify the value in dataframe d.

edited Jan 4 at 13:44

Ronak Shah

48k104269

asked Jan 4 at 13:31

DD chen

217

edited Jan 4 at 13:44

Ronak Shah

48k104269

asked Jan 4 at 13:31

DD chen

217

edited Jan 4 at 13:44

Ronak Shah

48k104269

edited Jan 4 at 13:44

Ronak Shah

48k104269

edited Jan 4 at 13:44

Ronak Shah

48k104269

asked Jan 4 at 13:31

DD chen

217

asked Jan 4 at 13:31

DD chen

217

asked Jan 4 at 13:31

DD chen

217

I think you have a typo in the third row of your expected x output.

– Tim Biegeleisen
Jan 4 at 13:35

add a comment |

I think you have a typo in the third row of your expected x output.

– Tim Biegeleisen
Jan 4 at 13:35

I think you have a typo in the third row of your expected x output.

– Tim Biegeleisen
Jan 4 at 13:35

add a comment |

3 Answers
3

active

oldest

votes

With dplyr and case_when:

library(dplyr)

d <- d %<% 

mutate(target = 3,

       x = case_when(x < target ~ 0,

                     T ~ x),

       y = case_when(y > target ~ 9,

                     T ~ y))

With base r:

d$x[d$x < target,] <- 0

d$y[d$y > target,] <- 9

edited Jan 4 at 13:42

answered Jan 4 at 13:37

Tom Haddow

17510

thank you for your response, i have a second question, if the target is a vector (same length as nrow(d) ). that means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:48

If you have a vector containing all of the targets e.g. targets <- c(1, 2, 3) then yes you can mutate this onto your data frame and then use case_when to compare with the targets column and change your x/y columns accordingly.

– Tom Haddow
Jan 4 at 13:52

Essentially mutate will compare row by row but for this to work your target vector needs to be a column in your data frame. If you already have your target as a vector the dplyr example I posted should work except you would change the line mutate(target = 3, to be mutate(target = target, so that your target vector is now the "target" column of your data frame.

– Tom Haddow
Jan 4 at 13:58

1

thanks a lot , i try with base r and that works! and if i want the target become the value that i want affect to (if the condition is true), i just run: d$y[d$y > target,] <- target[d$y > target,]

– DD chen
Jan 4 at 14:20

add a comment |

Multiple ways to do this. One way using double replace

replace(replace(d, d > target, 9), d < target, 0)



#  x y

#1 0 9

#2 0 9

#3 3 9

This logic can also be used in dplyr chain

library(dplyr)

d %>% mutate_all(funs(replace(replace(., . > target, 9), . < target, 0)))

answered Jan 4 at 13:35

Ronak Shah

48k104269

thank you for your response, but if the target is also a vector (same length as nrow(d). which means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:46

@DDchen nope, If you have target as vector I don't think mutate would work as it is as shown in the answer but replace would still work. Please check it on your real data.

– Ronak Shah
Jan 4 at 13:53

thank you @Ronak Shah !

– DD chen
Jan 4 at 14:22

@DDchen you are welcome. Just FYI, you can select only one answer as accepted.

– Ronak Shah
Jan 4 at 14:26

add a comment |

A straighforward way only using R base.

# Add object.

target <- 3



# Create dataframe.

df1 <- data.frame(x = c(1, 2, 3),y=c(4, 5, 6))



# Copy dataframe [df1] to new dataframe [ðf2].

# Preserve old dataframe [df1] for comparison.

df2 <- df1



# Insert digits.

df2$x[df2$x < target] <- 0

df2$y[df2$y > target] <- 9

answered Jan 5 at 15:27

Toolbox

672312

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

With dplyr and case_when:

library(dplyr)

d <- d %<% 

mutate(target = 3,

       x = case_when(x < target ~ 0,

                     T ~ x),

       y = case_when(y > target ~ 9,

                     T ~ y))

With base r:

d$x[d$x < target,] <- 0

d$y[d$y > target,] <- 9

edited Jan 4 at 13:42

answered Jan 4 at 13:37

Tom Haddow

17510

thank you for your response, i have a second question, if the target is a vector (same length as nrow(d) ). that means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:48

If you have a vector containing all of the targets e.g. targets <- c(1, 2, 3) then yes you can mutate this onto your data frame and then use case_when to compare with the targets column and change your x/y columns accordingly.

– Tom Haddow
Jan 4 at 13:52

Essentially mutate will compare row by row but for this to work your target vector needs to be a column in your data frame. If you already have your target as a vector the dplyr example I posted should work except you would change the line mutate(target = 3, to be mutate(target = target, so that your target vector is now the "target" column of your data frame.

– Tom Haddow
Jan 4 at 13:58

1

thanks a lot , i try with base r and that works! and if i want the target become the value that i want affect to (if the condition is true), i just run: d$y[d$y > target,] <- target[d$y > target,]

– DD chen
Jan 4 at 14:20

add a comment |

With dplyr and case_when:

library(dplyr)

d <- d %<% 

mutate(target = 3,

       x = case_when(x < target ~ 0,

                     T ~ x),

       y = case_when(y > target ~ 9,

                     T ~ y))

With base r:

d$x[d$x < target,] <- 0

d$y[d$y > target,] <- 9

edited Jan 4 at 13:42

answered Jan 4 at 13:37

Tom Haddow

17510

thank you for your response, i have a second question, if the target is a vector (same length as nrow(d) ). that means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:48

If you have a vector containing all of the targets e.g. targets <- c(1, 2, 3) then yes you can mutate this onto your data frame and then use case_when to compare with the targets column and change your x/y columns accordingly.

– Tom Haddow
Jan 4 at 13:52

Essentially mutate will compare row by row but for this to work your target vector needs to be a column in your data frame. If you already have your target as a vector the dplyr example I posted should work except you would change the line mutate(target = 3, to be mutate(target = target, so that your target vector is now the "target" column of your data frame.

– Tom Haddow
Jan 4 at 13:58

1

thanks a lot , i try with base r and that works! and if i want the target become the value that i want affect to (if the condition is true), i just run: d$y[d$y > target,] <- target[d$y > target,]

– DD chen
Jan 4 at 14:20

add a comment |

With dplyr and case_when:

library(dplyr)

d <- d %<% 

mutate(target = 3,

       x = case_when(x < target ~ 0,

                     T ~ x),

       y = case_when(y > target ~ 9,

                     T ~ y))

With base r:

d$x[d$x < target,] <- 0

d$y[d$y > target,] <- 9

edited Jan 4 at 13:42

answered Jan 4 at 13:37

Tom Haddow

17510

With dplyr and case_when:

library(dplyr)

d <- d %<% 

mutate(target = 3,

       x = case_when(x < target ~ 0,

                     T ~ x),

       y = case_when(y > target ~ 9,

                     T ~ y))

With base r:

d$x[d$x < target,] <- 0

d$y[d$y > target,] <- 9

edited Jan 4 at 13:42

answered Jan 4 at 13:37

Tom Haddow

17510

edited Jan 4 at 13:42

answered Jan 4 at 13:37

Tom Haddow

17510

answered Jan 4 at 13:37

Tom Haddow

17510

answered Jan 4 at 13:37

Tom Haddow

17510

thank you for your response, i have a second question, if the target is a vector (same length as nrow(d) ). that means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:48

If you have a vector containing all of the targets e.g. targets <- c(1, 2, 3) then yes you can mutate this onto your data frame and then use case_when to compare with the targets column and change your x/y columns accordingly.

– Tom Haddow
Jan 4 at 13:52

Essentially mutate will compare row by row but for this to work your target vector needs to be a column in your data frame. If you already have your target as a vector the dplyr example I posted should work except you would change the line mutate(target = 3, to be mutate(target = target, so that your target vector is now the "target" column of your data frame.

– Tom Haddow
Jan 4 at 13:58

1

thanks a lot , i try with base r and that works! and if i want the target become the value that i want affect to (if the condition is true), i just run: d$y[d$y > target,] <- target[d$y > target,]

– DD chen
Jan 4 at 14:20

add a comment |

thank you for your response, i have a second question, if the target is a vector (same length as nrow(d) ). that means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:48

If you have a vector containing all of the targets e.g. targets <- c(1, 2, 3) then yes you can mutate this onto your data frame and then use case_when to compare with the targets column and change your x/y columns accordingly.

– Tom Haddow
Jan 4 at 13:52

Essentially mutate will compare row by row but for this to work your target vector needs to be a column in your data frame. If you already have your target as a vector the dplyr example I posted should work except you would change the line mutate(target = 3, to be mutate(target = target, so that your target vector is now the "target" column of your data frame.

– Tom Haddow
Jan 4 at 13:58

1

thanks a lot , i try with base r and that works! and if i want the target become the value that i want affect to (if the condition is true), i just run: d$y[d$y > target,] <- target[d$y > target,]

– DD chen
Jan 4 at 14:20

thank you for your response, i have a second question, if the target is a vector (same length as nrow(d) ). that means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:48

If you have a vector containing all of the targets e.g. targets <- c(1, 2, 3) then yes you can mutate this onto your data frame and then use case_when to compare with the targets column and change your x/y columns accordingly.

– Tom Haddow
Jan 4 at 13:52

Essentially mutate will compare row by row but for this to work your target vector needs to be a column in your data frame. If you already have your target as a vector the dplyr example I posted should work except you would change the line mutate(target = 3, to be mutate(target = target, so that your target vector is now the "target" column of your data frame.

– Tom Haddow
Jan 4 at 13:58

thanks a lot , i try with base r and that works! and if i want the target become the value that i want affect to (if the condition is true), i just run: d$y[d$y > target,] <- target[d$y > target,]

– DD chen
Jan 4 at 14:20

add a comment |

Multiple ways to do this. One way using double replace

replace(replace(d, d > target, 9), d < target, 0)



#  x y

#1 0 9

#2 0 9

#3 3 9

This logic can also be used in dplyr chain

library(dplyr)

d %>% mutate_all(funs(replace(replace(., . > target, 9), . < target, 0)))

answered Jan 4 at 13:35

Ronak Shah

48k104269

thank you for your response, but if the target is also a vector (same length as nrow(d). which means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:46

@DDchen nope, If you have target as vector I don't think mutate would work as it is as shown in the answer but replace would still work. Please check it on your real data.

– Ronak Shah
Jan 4 at 13:53

thank you @Ronak Shah !

– DD chen
Jan 4 at 14:22

@DDchen you are welcome. Just FYI, you can select only one answer as accepted.

– Ronak Shah
Jan 4 at 14:26

add a comment |

Multiple ways to do this. One way using double replace

replace(replace(d, d > target, 9), d < target, 0)



#  x y

#1 0 9

#2 0 9

#3 3 9

This logic can also be used in dplyr chain

library(dplyr)

d %>% mutate_all(funs(replace(replace(., . > target, 9), . < target, 0)))

answered Jan 4 at 13:35

Ronak Shah

48k104269

thank you for your response, but if the target is also a vector (same length as nrow(d). which means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:46

@DDchen nope, If you have target as vector I don't think mutate would work as it is as shown in the answer but replace would still work. Please check it on your real data.

– Ronak Shah
Jan 4 at 13:53

thank you @Ronak Shah !

– DD chen
Jan 4 at 14:22

@DDchen you are welcome. Just FYI, you can select only one answer as accepted.

– Ronak Shah
Jan 4 at 14:26

add a comment |

Multiple ways to do this. One way using double replace

replace(replace(d, d > target, 9), d < target, 0)



#  x y

#1 0 9

#2 0 9

#3 3 9

This logic can also be used in dplyr chain

library(dplyr)

d %>% mutate_all(funs(replace(replace(., . > target, 9), . < target, 0)))

answered Jan 4 at 13:35

Ronak Shah

48k104269

Multiple ways to do this. One way using double replace

replace(replace(d, d > target, 9), d < target, 0)



#  x y

#1 0 9

#2 0 9

#3 3 9

This logic can also be used in dplyr chain

library(dplyr)

d %>% mutate_all(funs(replace(replace(., . > target, 9), . < target, 0)))

answered Jan 4 at 13:35

Ronak Shah

48k104269

answered Jan 4 at 13:35

Ronak Shah

48k104269

answered Jan 4 at 13:35

Ronak Shah

48k104269

answered Jan 4 at 13:35

Ronak Shah

48k104269

thank you for your response, but if the target is also a vector (same length as nrow(d). which means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:46

@DDchen nope, If you have target as vector I don't think mutate would work as it is as shown in the answer but replace would still work. Please check it on your real data.

– Ronak Shah
Jan 4 at 13:53

thank you @Ronak Shah !

– DD chen
Jan 4 at 14:22

@DDchen you are welcome. Just FYI, you can select only one answer as accepted.

– Ronak Shah
Jan 4 at 14:26

add a comment |

thank you for your response, but if the target is also a vector (same length as nrow(d). which means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:46

@DDchen nope, If you have target as vector I don't think mutate would work as it is as shown in the answer but replace would still work. Please check it on your real data.

– Ronak Shah
Jan 4 at 13:53

thank you @Ronak Shah !

– DD chen
Jan 4 at 14:22

@DDchen you are welcome. Just FYI, you can select only one answer as accepted.

– Ronak Shah
Jan 4 at 14:26

thank you for your response, but if the target is also a vector (same length as nrow(d). which means the target is different for each row. do the function mutate still work?

– DD chen
Jan 4 at 13:46

@DDchen nope, If you have target as vector I don't think mutate would work as it is as shown in the answer but replace would still work. Please check it on your real data.

– Ronak Shah
Jan 4 at 13:53

thank you @Ronak Shah !

– DD chen
Jan 4 at 14:22

@DDchen you are welcome. Just FYI, you can select only one answer as accepted.

– Ronak Shah
Jan 4 at 14:26

add a comment |

A straighforward way only using R base.

# Add object.

target <- 3



# Create dataframe.

df1 <- data.frame(x = c(1, 2, 3),y=c(4, 5, 6))



# Copy dataframe [df1] to new dataframe [ðf2].

# Preserve old dataframe [df1] for comparison.

df2 <- df1



# Insert digits.

df2$x[df2$x < target] <- 0

df2$y[df2$y > target] <- 9

answered Jan 5 at 15:27

Toolbox

672312

add a comment |

A straighforward way only using R base.

# Add object.

target <- 3



# Create dataframe.

df1 <- data.frame(x = c(1, 2, 3),y=c(4, 5, 6))



# Copy dataframe [df1] to new dataframe [ðf2].

# Preserve old dataframe [df1] for comparison.

df2 <- df1



# Insert digits.

df2$x[df2$x < target] <- 0

df2$y[df2$y > target] <- 9

answered Jan 5 at 15:27

Toolbox

672312

add a comment |

A straighforward way only using R base.

# Add object.

target <- 3



# Create dataframe.

df1 <- data.frame(x = c(1, 2, 3),y=c(4, 5, 6))



# Copy dataframe [df1] to new dataframe [ðf2].

# Preserve old dataframe [df1] for comparison.

df2 <- df1



# Insert digits.

df2$x[df2$x < target] <- 0

df2$y[df2$y > target] <- 9

answered Jan 5 at 15:27

Toolbox

672312

A straighforward way only using R base.

# Add object.

target <- 3



# Create dataframe.

df1 <- data.frame(x = c(1, 2, 3),y=c(4, 5, 6))



# Copy dataframe [df1] to new dataframe [ðf2].

# Preserve old dataframe [df1] for comparison.

df2 <- df1



# Insert digits.

df2$x[df2$x < target] <- 0

df2$y[df2$y > target] <- 9

answered Jan 5 at 15:27

Toolbox

672312

answered Jan 5 at 15:27

Toolbox

672312

answered Jan 5 at 15:27

Toolbox

672312

answered Jan 5 at 15:27

Toolbox

672312

add a comment |

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