mutate_at function cancels the previous mutate_at

0

I have created a dataframe S by merging two dataframes innov2015 and innov2017 by a unique identifying column.
Some cases in innov2015 are not included in innov2017 and vice versa, so there are NA entries for half of the variables in S for some of the cases.

I want to calculate p = (p_2015+p_2017)/2 , however, when there is an NA entry for p_2015 I want p = p_2017 and vice versa.

I have tried to do this with:

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2017), p_2015,(p_2015+p_2017)/2))) %>%  

If I run

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

p takes the desired value.

when I run both mutate_at() statements

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2017), p_2015,(p_2015+p_2017)/2))) %>% 

the second mutate_at() statement produces the desired values, however it undoes the first mutate_at() statement and where p had taken the correct value, there is now NA

What do I need to do to make both mutate_at() statements work without cancelling the previous one?

r condition mutate

share|improve this question

asked Jan 2 at 12:06

LauraLaura

596

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Why don't you just take the mean of both variables? If one is missing, it takes the mean of 1 value, which is exactly what you want in case not both values are available? Do not forget to specify the na.rm argument in the mean() function to TRUE. S %>% mutate(p = mean(c(p_2015,p_2017), na.rm = T))
  
  – Lennyy
  Jan 2 at 12:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for the work around, do you know have an explanation for why my code doesn't work? EDIT: the code you have provided inputs the value of p as the mean of the entire columns p_2015 and p_2017 rather than the mean of the values in the corresponding row
  
  – Laura
  Jan 2 at 12:15
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Sorry I meant the rowMeans() function above, but can't edit that comment anymore. Anyway, I think your attempt is more of a workaround than my suggestion. Something like: S$p <- rowMeans(S[,c("p_2015", "p_2017")], na.rm = T) with just base R would work. The answer below beat me to it with regard to explaining why your attempt did not have the desired result.
  
  – Lennyy
  Jan 2 at 12:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks that works
  
  – Laura
  Jan 2 at 12:23
  

  
  
  
  

add a comment | 

0

I have created a dataframe S by merging two dataframes innov2015 and innov2017 by a unique identifying column.
Some cases in innov2015 are not included in innov2017 and vice versa, so there are NA entries for half of the variables in S for some of the cases.

I want to calculate p = (p_2015+p_2017)/2 , however, when there is an NA entry for p_2015 I want p = p_2017 and vice versa.

I have tried to do this with:

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2017), p_2015,(p_2015+p_2017)/2))) %>%  

If I run

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

p takes the desired value.

when I run both mutate_at() statements

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2017), p_2015,(p_2015+p_2017)/2))) %>% 

the second mutate_at() statement produces the desired values, however it undoes the first mutate_at() statement and where p had taken the correct value, there is now NA

What do I need to do to make both mutate_at() statements work without cancelling the previous one?

r condition mutate

share|improve this question

asked Jan 2 at 12:06

LauraLaura

596

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Why don't you just take the mean of both variables? If one is missing, it takes the mean of 1 value, which is exactly what you want in case not both values are available? Do not forget to specify the na.rm argument in the mean() function to TRUE. S %>% mutate(p = mean(c(p_2015,p_2017), na.rm = T))
  
  – Lennyy
  Jan 2 at 12:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for the work around, do you know have an explanation for why my code doesn't work? EDIT: the code you have provided inputs the value of p as the mean of the entire columns p_2015 and p_2017 rather than the mean of the values in the corresponding row
  
  – Laura
  Jan 2 at 12:15
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Sorry I meant the rowMeans() function above, but can't edit that comment anymore. Anyway, I think your attempt is more of a workaround than my suggestion. Something like: S$p <- rowMeans(S[,c("p_2015", "p_2017")], na.rm = T) with just base R would work. The answer below beat me to it with regard to explaining why your attempt did not have the desired result.
  
  – Lennyy
  Jan 2 at 12:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks that works
  
  – Laura
  Jan 2 at 12:23
  

  
  
  
  

add a comment | 

0

0

0

I have created a dataframe S by merging two dataframes innov2015 and innov2017 by a unique identifying column.
Some cases in innov2015 are not included in innov2017 and vice versa, so there are NA entries for half of the variables in S for some of the cases.

I want to calculate p = (p_2015+p_2017)/2 , however, when there is an NA entry for p_2015 I want p = p_2017 and vice versa.

I have tried to do this with:

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2017), p_2015,(p_2015+p_2017)/2))) %>%  

If I run

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

p takes the desired value.

when I run both mutate_at() statements

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2017), p_2015,(p_2015+p_2017)/2))) %>% 

the second mutate_at() statement produces the desired values, however it undoes the first mutate_at() statement and where p had taken the correct value, there is now NA

What do I need to do to make both mutate_at() statements work without cancelling the previous one?

r condition mutate

share|improve this question

asked Jan 2 at 12:06

LauraLaura

596

I have created a dataframe S by merging two dataframes innov2015 and innov2017 by a unique identifying column.
Some cases in innov2015 are not included in innov2017 and vice versa, so there are NA entries for half of the variables in S for some of the cases.

I want to calculate p = (p_2015+p_2017)/2 , however, when there is an NA entry for p_2015 I want p = p_2017 and vice versa.

I have tried to do this with:

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2017), p_2015,(p_2015+p_2017)/2))) %>%  

If I run

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

p takes the desired value.

when I run both mutate_at() statements

    S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = 0) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2015), p_2017,(p_2015+p_2017)/2))) %>%

  mutate_at(vars(p), funs(ifelse(is.na(smalln_2017), p_2015,(p_2015+p_2017)/2))) %>% 

the second mutate_at() statement produces the desired values, however it undoes the first mutate_at() statement and where p had taken the correct value, there is now NA

What do I need to do to make both mutate_at() statements work without cancelling the previous one?

r condition mutate

r condition mutate

share|improve this question

asked Jan 2 at 12:06

LauraLaura

596

share|improve this question

asked Jan 2 at 12:06

LauraLaura

596

share|improve this question

share|improve this question

asked Jan 2 at 12:06

LauraLaura

596

asked Jan 2 at 12:06

LauraLaura

596

asked Jan 2 at 12:06

LauraLaura

596

596

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Why don't you just take the mean of both variables? If one is missing, it takes the mean of 1 value, which is exactly what you want in case not both values are available? Do not forget to specify the na.rm argument in the mean() function to TRUE. S %>% mutate(p = mean(c(p_2015,p_2017), na.rm = T))
  
  – Lennyy
  Jan 2 at 12:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for the work around, do you know have an explanation for why my code doesn't work? EDIT: the code you have provided inputs the value of p as the mean of the entire columns p_2015 and p_2017 rather than the mean of the values in the corresponding row
  
  – Laura
  Jan 2 at 12:15
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Sorry I meant the rowMeans() function above, but can't edit that comment anymore. Anyway, I think your attempt is more of a workaround than my suggestion. Something like: S$p <- rowMeans(S[,c("p_2015", "p_2017")], na.rm = T) with just base R would work. The answer below beat me to it with regard to explaining why your attempt did not have the desired result.
  
  – Lennyy
  Jan 2 at 12:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks that works
  
  – Laura
  Jan 2 at 12:23
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Why don't you just take the mean of both variables? If one is missing, it takes the mean of 1 value, which is exactly what you want in case not both values are available? Do not forget to specify the na.rm argument in the mean() function to TRUE. S %>% mutate(p = mean(c(p_2015,p_2017), na.rm = T))
  
  – Lennyy
  Jan 2 at 12:09
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks for the work around, do you know have an explanation for why my code doesn't work? EDIT: the code you have provided inputs the value of p as the mean of the entire columns p_2015 and p_2017 rather than the mean of the values in the corresponding row
  
  – Laura
  Jan 2 at 12:15
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Sorry I meant the rowMeans() function above, but can't edit that comment anymore. Anyway, I think your attempt is more of a workaround than my suggestion. Something like: S$p <- rowMeans(S[,c("p_2015", "p_2017")], na.rm = T) with just base R would work. The answer below beat me to it with regard to explaining why your attempt did not have the desired result.
  
  – Lennyy
  Jan 2 at 12:18
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks that works
  
  – Laura
  Jan 2 at 12:23
  

  
  
  
  

1

1

Why don't you just take the mean of both variables? If one is missing, it takes the mean of 1 value, which is exactly what you want in case not both values are available? Do not forget to specify the na.rm argument in the mean() function to TRUE. S %>% mutate(p = mean(c(p_2015,p_2017), na.rm = T))

– Lennyy
Jan 2 at 12:09

Why don't you just take the mean of both variables? If one is missing, it takes the mean of 1 value, which is exactly what you want in case not both values are available? Do not forget to specify the na.rm argument in the mean() function to TRUE. S %>% mutate(p = mean(c(p_2015,p_2017), na.rm = T))

– Lennyy
Jan 2 at 12:09

Thanks for the work around, do you know have an explanation for why my code doesn't work? EDIT: the code you have provided inputs the value of p as the mean of the entire columns p_2015 and p_2017 rather than the mean of the values in the corresponding row

– Laura
Jan 2 at 12:15

Thanks for the work around, do you know have an explanation for why my code doesn't work? EDIT: the code you have provided inputs the value of p as the mean of the entire columns p_2015 and p_2017 rather than the mean of the values in the corresponding row

– Laura
Jan 2 at 12:15

2

2

Sorry I meant the rowMeans() function above, but can't edit that comment anymore. Anyway, I think your attempt is more of a workaround than my suggestion. Something like: S$p <- rowMeans(S[,c("p_2015", "p_2017")], na.rm = T) with just base R would work. The answer below beat me to it with regard to explaining why your attempt did not have the desired result.

– Lennyy
Jan 2 at 12:18

Sorry I meant the rowMeans() function above, but can't edit that comment anymore. Anyway, I think your attempt is more of a workaround than my suggestion. Something like: S$p <- rowMeans(S[,c("p_2015", "p_2017")], na.rm = T) with just base R would work. The answer below beat me to it with regard to explaining why your attempt did not have the desired result.

– Lennyy
Jan 2 at 12:18

Thanks that works

– Laura
Jan 2 at 12:23

Thanks that works

– Laura
Jan 2 at 12:23

add a comment | 

1 Answer
1

active

oldest

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1

These two mutates conflict. You are fully re-defining "p" in each of them, since the value of "p" from the first call is never re-used in the second. @Lennyy's comment will get the job done, but if you want to keep this operation within the tidyverse, you might have better luck using case_when. Your example is not fully reproducible, so the following is a guess as to how it should work:

S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = case_when(

    is.na(smalln_2015) ~ smalln_2017,

    is.na(smalln_2017) ~ smalln_2015,

    TRUE ~ (smalln_2015 + smalln_2017) / 2

  ))

share|improve this answer

edited Jan 2 at 12:26

answered Jan 2 at 12:18

jdobresjdobres

5,0231623

add a comment | 

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These two mutates conflict. You are fully re-defining "p" in each of them, since the value of "p" from the first call is never re-used in the second. @Lennyy's comment will get the job done, but if you want to keep this operation within the tidyverse, you might have better luck using case_when. Your example is not fully reproducible, so the following is a guess as to how it should work:

S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = case_when(

    is.na(smalln_2015) ~ smalln_2017,

    is.na(smalln_2017) ~ smalln_2015,

    TRUE ~ (smalln_2015 + smalln_2017) / 2

  ))

share|improve this answer

edited Jan 2 at 12:26

answered Jan 2 at 12:18

jdobresjdobres

5,0231623

add a comment | 

1

These two mutates conflict. You are fully re-defining "p" in each of them, since the value of "p" from the first call is never re-used in the second. @Lennyy's comment will get the job done, but if you want to keep this operation within the tidyverse, you might have better luck using case_when. Your example is not fully reproducible, so the following is a guess as to how it should work:

S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = case_when(

    is.na(smalln_2015) ~ smalln_2017,

    is.na(smalln_2017) ~ smalln_2015,

    TRUE ~ (smalln_2015 + smalln_2017) / 2

  ))

share|improve this answer

edited Jan 2 at 12:26

answered Jan 2 at 12:18

jdobresjdobres

5,0231623

add a comment | 

1

1

1

These two mutates conflict. You are fully re-defining "p" in each of them, since the value of "p" from the first call is never re-used in the second. @Lennyy's comment will get the job done, but if you want to keep this operation within the tidyverse, you might have better luck using case_when. Your example is not fully reproducible, so the following is a guess as to how it should work:

S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = case_when(

    is.na(smalln_2015) ~ smalln_2017,

    is.na(smalln_2017) ~ smalln_2015,

    TRUE ~ (smalln_2015 + smalln_2017) / 2

  ))

share|improve this answer

edited Jan 2 at 12:26

answered Jan 2 at 12:18

jdobresjdobres

5,0231623

These two mutates conflict. You are fully re-defining "p" in each of them, since the value of "p" from the first call is never re-used in the second. @Lennyy's comment will get the job done, but if you want to keep this operation within the tidyverse, you might have better luck using case_when. Your example is not fully reproducible, so the following is a guess as to how it should work:

S <- merge(x = innov_2015_2, y = innov_2017_2, by = "cell_no", all = TRUE) %>%

  mutate(p = case_when(

    is.na(smalln_2015) ~ smalln_2017,

    is.na(smalln_2017) ~ smalln_2015,

    TRUE ~ (smalln_2015 + smalln_2017) / 2

  ))

share|improve this answer

edited Jan 2 at 12:26

answered Jan 2 at 12:18

jdobresjdobres

5,0231623

share|improve this answer

share|improve this answer

edited Jan 2 at 12:26

edited Jan 2 at 12:26

edited Jan 2 at 12:26

answered Jan 2 at 12:18

jdobresjdobres

5,0231623

answered Jan 2 at 12:18

jdobresjdobres

5,0231623

answered Jan 2 at 12:18

jdobresjdobres

5,0231623

5,0231623

add a comment | 

add a comment | 

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