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Is there a better solution? Receiving “longer object length is not a multiple of shorter object length”

Longer object length is not a multiple of shorter object length?Strange thing happening: longer object length is not a multiple of shorter object lengthlonger object length is not a multiple of shorter object lengthWarning: longer object length is not a multiple of shorter object lengthR warning - longer object length is not a multiple of shorter object lengthCorrect result but warnings that longer object length is not a multiple of shorter object lengthWarning: longer object length is not a multiple of shorter object length?R - longer object length is not a multiple of shorter object lengthlonger object length is not a multiple of shorter object length in r“longer object length is not a multiple of shorter object length”

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

0

This is my example.

user_id <- sample(seq(1,100),5000, TRUE)
friend_id <- sample(seq(1,100),5000, TRUE)
friends <- data.frame(user_id, friend_id) 
friends <- friends %>% 
 filter(!user_id == friend_id)
friends <- friends %>% arrange(user_id) %>% distinct()

user_id <- sample(seq(1,100),10000, TRUE)
page_id <- sample(seq(1000,2000),10000, TRUE)
pages <- data.frame(user_id, page_id)
pages <- arrange(pages, user_id) %>% distinct()

popular <- friends %>%
 left_join(pages, by = c("friend_id" = "user_id")) %>%
 group_by(user_id, page_id) %>%
 summarize(likes = n()) %>%
 arrange(-likes) %>%
 filter(!page_id %in% pages[pages$user_id == user_id,]$page_id)

My goal is to count the number of likes for each of the pages that a user's friend has liked. The last step is giving me this warning:

50: In pages$user_id == user_id : longer object length is not a
multiple of shorter object length

My goal in the last step is to filter out any page that the user has liked.

1) If I group by a column and then apply filter, will it apply to each of the grouped data frames separately? In other words, is it like having a for loop that says for (group in tbl) apply filter?

2) Will user_id give me the user_id according to each group? I guess this is an extension of 1.

3) I think it gives me the warning since pages$user_id is long and user_id is just one value. Is there a better solution or a more appropriate solution?

r

share|improve this question

edited Mar 27 at 4:21

Cauder

asked Mar 27 at 4:03

CauderCauder

1921111 bronze badges

add a comment | 

0

This is my example.

user_id <- sample(seq(1,100),5000, TRUE)
friend_id <- sample(seq(1,100),5000, TRUE)
friends <- data.frame(user_id, friend_id) 
friends <- friends %>% 
 filter(!user_id == friend_id)
friends <- friends %>% arrange(user_id) %>% distinct()

user_id <- sample(seq(1,100),10000, TRUE)
page_id <- sample(seq(1000,2000),10000, TRUE)
pages <- data.frame(user_id, page_id)
pages <- arrange(pages, user_id) %>% distinct()

popular <- friends %>%
 left_join(pages, by = c("friend_id" = "user_id")) %>%
 group_by(user_id, page_id) %>%
 summarize(likes = n()) %>%
 arrange(-likes) %>%
 filter(!page_id %in% pages[pages$user_id == user_id,]$page_id)

My goal is to count the number of likes for each of the pages that a user's friend has liked. The last step is giving me this warning:

50: In pages$user_id == user_id : longer object length is not a
multiple of shorter object length

My goal in the last step is to filter out any page that the user has liked.

1) If I group by a column and then apply filter, will it apply to each of the grouped data frames separately? In other words, is it like having a for loop that says for (group in tbl) apply filter?

2) Will user_id give me the user_id according to each group? I guess this is an extension of 1.

3) I think it gives me the warning since pages$user_id is long and user_id is just one value. Is there a better solution or a more appropriate solution?

r

share|improve this question

edited Mar 27 at 4:21

Cauder

asked Mar 27 at 4:03

CauderCauder

1921111 bronze badges

add a comment | 

This is my example.

user_id <- sample(seq(1,100),5000, TRUE)
friend_id <- sample(seq(1,100),5000, TRUE)
friends <- data.frame(user_id, friend_id) 
friends <- friends %>% 
 filter(!user_id == friend_id)
friends <- friends %>% arrange(user_id) %>% distinct()

user_id <- sample(seq(1,100),10000, TRUE)
page_id <- sample(seq(1000,2000),10000, TRUE)
pages <- data.frame(user_id, page_id)
pages <- arrange(pages, user_id) %>% distinct()

popular <- friends %>%
 left_join(pages, by = c("friend_id" = "user_id")) %>%
 group_by(user_id, page_id) %>%
 summarize(likes = n()) %>%
 arrange(-likes) %>%
 filter(!page_id %in% pages[pages$user_id == user_id,]$page_id)

My goal is to count the number of likes for each of the pages that a user's friend has liked. The last step is giving me this warning:

50: In pages$user_id == user_id : longer object length is not a
multiple of shorter object length

My goal in the last step is to filter out any page that the user has liked.

1) If I group by a column and then apply filter, will it apply to each of the grouped data frames separately? In other words, is it like having a for loop that says for (group in tbl) apply filter?

2) Will user_id give me the user_id according to each group? I guess this is an extension of 1.

3) I think it gives me the warning since pages$user_id is long and user_id is just one value. Is there a better solution or a more appropriate solution?

r

share|improve this question

edited Mar 27 at 4:21

Cauder

asked Mar 27 at 4:03

CauderCauder

1921111 bronze badges

user_id <- sample(seq(1,100),5000, TRUE)
friend_id <- sample(seq(1,100),5000, TRUE)
friends <- data.frame(user_id, friend_id) 
friends <- friends %>% 
 filter(!user_id == friend_id)
friends <- friends %>% arrange(user_id) %>% distinct()

user_id <- sample(seq(1,100),10000, TRUE)
page_id <- sample(seq(1000,2000),10000, TRUE)
pages <- data.frame(user_id, page_id)
pages <- arrange(pages, user_id) %>% distinct()

popular <- friends %>%
 left_join(pages, by = c("friend_id" = "user_id")) %>%
 group_by(user_id, page_id) %>%
 summarize(likes = n()) %>%
 arrange(-likes) %>%
 filter(!page_id %in% pages[pages$user_id == user_id,]$page_id)

share|improve this question

edited Mar 27 at 4:21

Cauder

asked Mar 27 at 4:03

CauderCauder

1921111 bronze badges

share|improve this question

edited Mar 27 at 4:21

Cauder

asked Mar 27 at 4:03

CauderCauder

1921111 bronze badges

asked Mar 27 at 4:03

CauderCauder

1921111 bronze badges

asked Mar 27 at 4:03

CauderCauder

1921111 bronze badges

1 Answer
1

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0

Is this what you are looking for:

pages_agg <- pages %>%
 group_by(user_id) %>%
 summarise(likes = n()) 

left_join(friends, pages_agg, by = c("friend_id" = "user_id")) %>%
 head()
 user_id friend_id likes
1 1 44 107
2 1 76 90
3 1 36 116
4 1 4 110
5 1 57 93
6 1 32 96

share|improve this answer

answered Mar 27 at 6:00

SonnySonny

2,70511 gold badge66 silver badges1717 bronze badges

add a comment | 

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0

Is this what you are looking for:

pages_agg <- pages %>%
 group_by(user_id) %>%
 summarise(likes = n()) 

left_join(friends, pages_agg, by = c("friend_id" = "user_id")) %>%
 head()
 user_id friend_id likes
1 1 44 107
2 1 76 90
3 1 36 116
4 1 4 110
5 1 57 93
6 1 32 96

share|improve this answer

answered Mar 27 at 6:00

SonnySonny

2,70511 gold badge66 silver badges1717 bronze badges

add a comment | 

0

Is this what you are looking for:

pages_agg <- pages %>%
 group_by(user_id) %>%
 summarise(likes = n()) 

left_join(friends, pages_agg, by = c("friend_id" = "user_id")) %>%
 head()
 user_id friend_id likes
1 1 44 107
2 1 76 90
3 1 36 116
4 1 4 110
5 1 57 93
6 1 32 96

share|improve this answer

answered Mar 27 at 6:00

SonnySonny

2,70511 gold badge66 silver badges1717 bronze badges

add a comment | 

Is this what you are looking for:

pages_agg <- pages %>%
 group_by(user_id) %>%
 summarise(likes = n()) 

left_join(friends, pages_agg, by = c("friend_id" = "user_id")) %>%
 head()
 user_id friend_id likes
1 1 44 107
2 1 76 90
3 1 36 116
4 1 4 110
5 1 57 93
6 1 32 96

share|improve this answer

answered Mar 27 at 6:00

SonnySonny

2,70511 gold badge66 silver badges1717 bronze badges

pages_agg <- pages %>%
 group_by(user_id) %>%
 summarise(likes = n()) 

left_join(friends, pages_agg, by = c("friend_id" = "user_id")) %>%
 head()
 user_id friend_id likes
1 1 44 107
2 1 76 90
3 1 36 116
4 1 4 110
5 1 57 93
6 1 32 96

share|improve this answer

answered Mar 27 at 6:00

SonnySonny

2,70511 gold badge66 silver badges1717 bronze badges

answered Mar 27 at 6:00

SonnySonny

2,70511 gold badge66 silver badges1717 bronze badges

answered Mar 27 at 6:00

SonnySonny

2,70511 gold badge66 silver badges1717 bronze badges