Dataframe: Divide each group by a vector corresponding to each group in R?
0
I have a data frame like this:
df1 <- structure(list(user_id = c(1, 1, 1, 2, 2, 2, 3, 3, 3), param_a = c(123,
2.3, -9, 1, -0.03333, 4, -41, -12, 0.89)), .Names = c("user_id",
"param_a"), row.names = c(NA, -9L), class = c("tbl_df", "tbl",
"data.frame"))
and another dataframe of vectors:
df2 <- structure(list(user_id = c(1, 2, 3), param_b = c(34, 12, -0.89
)), .Names = c("user_id", "param_b"), row.names = c(NA, -3L), class = c("tbl_df",
"tbl", "data.frame"))
Now I want to divide each group in df1 by corresponding value in df2:
For example for a group of user 1 divide each row by param_b first vector:
user_id param_a
1 123/34
1 2.3/34
1 -9/34
2 1/12
2 -0.03333/12
2 4/12
....
for user 2 divide each row by param_b second vector.
Please advise how can I divide a grouped by user dataframe by a vector per each group?
P.S
If I have df1 extended to param_a, param_k, param_p
and df2 extended accordingly with param_b, param_l, param_r
How can I perform this kind of operation? @nicola suggested a very nice solution but I want to extend it.
r dataframe datatable
asked Nov 19 '18 at 13:59
SteveSSteveS
568211
add a comment |
0
I have a data frame like this:
df1 <- structure(list(user_id = c(1, 1, 1, 2, 2, 2, 3, 3, 3), param_a = c(123,
2.3, -9, 1, -0.03333, 4, -41, -12, 0.89)), .Names = c("user_id",
"param_a"), row.names = c(NA, -9L), class = c("tbl_df", "tbl",
"data.frame"))
and another dataframe of vectors:
df2 <- structure(list(user_id = c(1, 2, 3), param_b = c(34, 12, -0.89
)), .Names = c("user_id", "param_b"), row.names = c(NA, -3L), class = c("tbl_df",
"tbl", "data.frame"))
Now I want to divide each group in df1 by corresponding value in df2:
For example for a group of user 1 divide each row by param_b first vector:
user_id param_a
1 123/34
1 2.3/34
1 -9/34
2 1/12
2 -0.03333/12
2 4/12
....
for user 2 divide each row by param_b second vector.
Please advise how can I divide a grouped by user dataframe by a vector per each group?
P.S
If I have df1 extended to param_a, param_k, param_p
and df2 extended accordingly with param_b, param_l, param_r
How can I perform this kind of operation? @nicola suggested a very nice solution but I want to extend it.
r dataframe datatable
asked Nov 19 '18 at 13:59
SteveSSteveS
568211
2
Trydf1$param_a/df2$param_b[match(df1$user_id,df2$user_id)].
– nicola
Nov 19 '18 at 14:02
@nicola it doens't work
– SteveS
Nov 19 '18 at 14:35
@nicola, if I have param_a, param_k, param_p and param b is a vector of multiple numbers it doesn't work.
– SteveS
Nov 19 '18 at 14:37
It works on your example. You should state why it doesn't work and provide a better example. FWIW, these kinds of "it doesn't work", without nothing else are so frustrating and let me question why I'm still here on this site trying to help others.
– nicola
Nov 19 '18 at 14:41
@nicola, I am sorry. You are completely right. I just thought it's a generic way to do this, but when I did apply it on more variables it didn't work. Can you please advise? I will update my question shortly.
– SteveS
Nov 19 '18 at 14:43
add a comment |
0
0
0
I have a data frame like this:
df1 <- structure(list(user_id = c(1, 1, 1, 2, 2, 2, 3, 3, 3), param_a = c(123,
2.3, -9, 1, -0.03333, 4, -41, -12, 0.89)), .Names = c("user_id",
"param_a"), row.names = c(NA, -9L), class = c("tbl_df", "tbl",
"data.frame"))
and another dataframe of vectors:
df2 <- structure(list(user_id = c(1, 2, 3), param_b = c(34, 12, -0.89
)), .Names = c("user_id", "param_b"), row.names = c(NA, -3L), class = c("tbl_df",
"tbl", "data.frame"))
Now I want to divide each group in df1 by corresponding value in df2:
For example for a group of user 1 divide each row by param_b first vector:
user_id param_a
1 123/34
1 2.3/34
1 -9/34
2 1/12
2 -0.03333/12
2 4/12
....
for user 2 divide each row by param_b second vector.
Please advise how can I divide a grouped by user dataframe by a vector per each group?
P.S
If I have df1 extended to param_a, param_k, param_p
and df2 extended accordingly with param_b, param_l, param_r
How can I perform this kind of operation? @nicola suggested a very nice solution but I want to extend it.
r dataframe datatable
asked Nov 19 '18 at 13:59
SteveSSteveS
568211
I have a data frame like this:
df1 <- structure(list(user_id = c(1, 1, 1, 2, 2, 2, 3, 3, 3), param_a = c(123,
2.3, -9, 1, -0.03333, 4, -41, -12, 0.89)), .Names = c("user_id",
"param_a"), row.names = c(NA, -9L), class = c("tbl_df", "tbl",
"data.frame"))
and another dataframe of vectors:
df2 <- structure(list(user_id = c(1, 2, 3), param_b = c(34, 12, -0.89
)), .Names = c("user_id", "param_b"), row.names = c(NA, -3L), class = c("tbl_df",
"tbl", "data.frame"))
Now I want to divide each group in df1 by corresponding value in df2:
For example for a group of user 1 divide each row by param_b first vector:
user_id param_a
1 123/34
1 2.3/34
1 -9/34
2 1/12
2 -0.03333/12
2 4/12
....
for user 2 divide each row by param_b second vector.
Please advise how can I divide a grouped by user dataframe by a vector per each group?
P.S
If I have df1 extended to param_a, param_k, param_p
and df2 extended accordingly with param_b, param_l, param_r
How can I perform this kind of operation? @nicola suggested a very nice solution but I want to extend it.
r dataframe datatable
r dataframe datatable
asked Nov 19 '18 at 13:59
SteveSSteveS
568211
asked Nov 19 '18 at 13:59
SteveSSteveS
568211
edited Nov 19 '18 at 14:45
SteveS
asked Nov 19 '18 at 13:59
SteveSSteveS
568211
asked Nov 19 '18 at 13:59
SteveSSteveS
568211
asked Nov 19 '18 at 13:59
SteveSSteveS
568211
568211
2
Trydf1$param_a/df2$param_b[match(df1$user_id,df2$user_id)].
– nicola
Nov 19 '18 at 14:02
@nicola it doens't work
– SteveS
Nov 19 '18 at 14:35
@nicola, if I have param_a, param_k, param_p and param b is a vector of multiple numbers it doesn't work.
– SteveS
Nov 19 '18 at 14:37
It works on your example. You should state why it doesn't work and provide a better example. FWIW, these kinds of "it doesn't work", without nothing else are so frustrating and let me question why I'm still here on this site trying to help others.
– nicola
Nov 19 '18 at 14:41
@nicola, I am sorry. You are completely right. I just thought it's a generic way to do this, but when I did apply it on more variables it didn't work. Can you please advise? I will update my question shortly.
– SteveS
Nov 19 '18 at 14:43
add a comment |
2
Trydf1$param_a/df2$param_b[match(df1$user_id,df2$user_id)].
– nicola
Nov 19 '18 at 14:02
@nicola it doens't work
– SteveS
Nov 19 '18 at 14:35
@nicola, if I have param_a, param_k, param_p and param b is a vector of multiple numbers it doesn't work.
– SteveS
Nov 19 '18 at 14:37
It works on your example. You should state why it doesn't work and provide a better example. FWIW, these kinds of "it doesn't work", without nothing else are so frustrating and let me question why I'm still here on this site trying to help others.
– nicola
Nov 19 '18 at 14:41
@nicola, I am sorry. You are completely right. I just thought it's a generic way to do this, but when I did apply it on more variables it didn't work. Can you please advise? I will update my question shortly.
– SteveS
Nov 19 '18 at 14:43
2
2
Try
df1$param_a/df2$param_b[match(df1$user_id,df2$user_id)].– nicola
Nov 19 '18 at 14:02
Try
df1$param_a/df2$param_b[match(df1$user_id,df2$user_id)].– nicola
Nov 19 '18 at 14:02
@nicola it doens't work
– SteveS
Nov 19 '18 at 14:35
@nicola it doens't work
– SteveS
Nov 19 '18 at 14:35
@nicola, if I have param_a, param_k, param_p and param b is a vector of multiple numbers it doesn't work.
– SteveS
Nov 19 '18 at 14:37
@nicola, if I have param_a, param_k, param_p and param b is a vector of multiple numbers it doesn't work.
– SteveS
Nov 19 '18 at 14:37
It works on your example. You should state why it doesn't work and provide a better example. FWIW, these kinds of "it doesn't work", without nothing else are so frustrating and let me question why I'm still here on this site trying to help others.
– nicola
Nov 19 '18 at 14:41
It works on your example. You should state why it doesn't work and provide a better example. FWIW, these kinds of "it doesn't work", without nothing else are so frustrating and let me question why I'm still here on this site trying to help others.
– nicola
Nov 19 '18 at 14:41
@nicola, I am sorry. You are completely right. I just thought it's a generic way to do this, but when I did apply it on more variables it didn't work. Can you please advise? I will update my question shortly.
– SteveS
Nov 19 '18 at 14:43
@nicola, I am sorry. You are completely right. I just thought it's a generic way to do this, but when I did apply it on more variables it didn't work. Can you please advise? I will update my question shortly.
– SteveS
Nov 19 '18 at 14:43
add a comment |
1 Answer
1
active
oldest
votes
1
Something like this?
df1%>%
left_join(df2)%>%
mutate(result=param_a/param_b)
Joining, by = "user_id"
# A tibble: 9 x 4
user_id param_a param_b result
<dbl> <dbl> <dbl> <dbl>
1 1 123 34 3.62
2 1 2.3 34 0.0676
3 1 -9 34 -0.265
4 2 1 12 0.0833
5 2 -0.0333 12 -0.00278
6 2 4 12 0.333
7 3 -41 -0.89 46.1
8 3 -12 -0.89 13.5
9 3 0.89 -0.89 -1
answered Nov 19 '18 at 14:03
jyjekjyjek
1,504312
add a comment |
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1 Answer
1
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Something like this?
df1%>%
left_join(df2)%>%
mutate(result=param_a/param_b)
Joining, by = "user_id"
# A tibble: 9 x 4
user_id param_a param_b result
<dbl> <dbl> <dbl> <dbl>
1 1 123 34 3.62
2 1 2.3 34 0.0676
3 1 -9 34 -0.265
4 2 1 12 0.0833
5 2 -0.0333 12 -0.00278
6 2 4 12 0.333
7 3 -41 -0.89 46.1
8 3 -12 -0.89 13.5
9 3 0.89 -0.89 -1
answered Nov 19 '18 at 14:03
jyjekjyjek
1,504312
add a comment |
1
Something like this?
df1%>%
left_join(df2)%>%
mutate(result=param_a/param_b)
Joining, by = "user_id"
# A tibble: 9 x 4
user_id param_a param_b result
<dbl> <dbl> <dbl> <dbl>
1 1 123 34 3.62
2 1 2.3 34 0.0676
3 1 -9 34 -0.265
4 2 1 12 0.0833
5 2 -0.0333 12 -0.00278
6 2 4 12 0.333
7 3 -41 -0.89 46.1
8 3 -12 -0.89 13.5
9 3 0.89 -0.89 -1
answered Nov 19 '18 at 14:03
jyjekjyjek
1,504312
add a comment |
1
1
1
Something like this?
df1%>%
left_join(df2)%>%
mutate(result=param_a/param_b)
Joining, by = "user_id"
# A tibble: 9 x 4
user_id param_a param_b result
<dbl> <dbl> <dbl> <dbl>
1 1 123 34 3.62
2 1 2.3 34 0.0676
3 1 -9 34 -0.265
4 2 1 12 0.0833
5 2 -0.0333 12 -0.00278
6 2 4 12 0.333
7 3 -41 -0.89 46.1
8 3 -12 -0.89 13.5
9 3 0.89 -0.89 -1
answered Nov 19 '18 at 14:03
jyjekjyjek
1,504312
Something like this?
df1%>%
left_join(df2)%>%
mutate(result=param_a/param_b)
Joining, by = "user_id"
# A tibble: 9 x 4
user_id param_a param_b result
<dbl> <dbl> <dbl> <dbl>
1 1 123 34 3.62
2 1 2.3 34 0.0676
3 1 -9 34 -0.265
4 2 1 12 0.0833
5 2 -0.0333 12 -0.00278
6 2 4 12 0.333
7 3 -41 -0.89 46.1
8 3 -12 -0.89 13.5
9 3 0.89 -0.89 -1
answered Nov 19 '18 at 14:03
jyjekjyjek
1,504312
answered Nov 19 '18 at 14:03
jyjekjyjek
1,504312
answered Nov 19 '18 at 14:03
jyjekjyjek
1,504312
answered Nov 19 '18 at 14:03
jyjekjyjek
1,504312
1,504312
add a comment |
add a comment |
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2
Try
df1$param_a/df2$param_b[match(df1$user_id,df2$user_id)].– nicola
Nov 19 '18 at 14:02
@nicola it doens't work
– SteveS
Nov 19 '18 at 14:35
@nicola, if I have param_a, param_k, param_p and param b is a vector of multiple numbers it doesn't work.
– SteveS
Nov 19 '18 at 14:37
It works on your example. You should state why it doesn't work and provide a better example. FWIW, these kinds of "it doesn't work", without nothing else are so frustrating and let me question why I'm still here on this site trying to help others.
– nicola
Nov 19 '18 at 14:41
@nicola, I am sorry. You are completely right. I just thought it's a generic way to do this, but when I did apply it on more variables it didn't work. Can you please advise? I will update my question shortly.
– SteveS
Nov 19 '18 at 14:43