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”
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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;
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
asked Mar 27 at 4:03
CauderCauder
19211 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
asked Mar 27 at 4:03
CauderCauder
19211 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
asked Mar 27 at 4:03
CauderCauder
19211 bronze badges
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
r
asked Mar 27 at 4:03
CauderCauder
19211 bronze badges
asked Mar 27 at 4:03
CauderCauder
19211 bronze badges
edited Mar 27 at 4:21
Cauder
asked Mar 27 at 4:03
CauderCauder
19211 bronze badges
asked Mar 27 at 4:03
CauderCauder
19211 bronze badges
asked Mar 27 at 4:03
CauderCauder
19211 bronze badges
19211 bronze badges
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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
answered Mar 27 at 6:00
SonnySonny
2,7051 gold badge6 silver badges17 bronze badges
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
answered Mar 27 at 6:00
SonnySonny
2,7051 gold badge6 silver badges17 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
answered Mar 27 at 6:00
SonnySonny
2,7051 gold badge6 silver badges17 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
answered Mar 27 at 6:00
SonnySonny
2,7051 gold badge6 silver badges17 bronze badges
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
answered Mar 27 at 6:00
SonnySonny
2,7051 gold badge6 silver badges17 bronze badges
answered Mar 27 at 6:00
SonnySonny
2,7051 gold badge6 silver badges17 bronze badges
answered Mar 27 at 6:00
SonnySonny
2,7051 gold badge6 silver badges17 bronze badges
answered Mar 27 at 6:00
SonnySonny
2,7051 gold badge6 silver badges17 bronze badges
2,7051 gold badge6 silver badges17 bronze badges
add a comment |
add a comment |
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