Calculate row-wise proportions
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I have a data frame:
x <- data.frame(id = letters[1:3], val0 = 1:3, val1 = 4:6, val2 = 7:9)
# id val0 val1 val2
# 1 a 1 4 7
# 2 b 2 5 8
# 3 c 3 6 9
Within each row, I want to calculate the corresponding proportions (ratio) for each value. E.g. for the value in column "val0", I want to calculate row-wise val0 / (val0 + val1 + val2).
Desired output:
id val0 val1 val2
1 a 0.083 0.33 0.583
2 b 0.133 0.33 0.533
3 c 0.167 0.33 0.5
Can anyone tell me what's the best way to do this? Here it's just three columns, but there can be alot of columns.
r dataframe apply
edited Aug 24 '16 at 8:22
Henrik
42.6k994111
asked Apr 16 '13 at 9:02
Rachit AgrawalRachit Agrawal
1,48452148
add a comment |
I have a data frame:
x <- data.frame(id = letters[1:3], val0 = 1:3, val1 = 4:6, val2 = 7:9)
# id val0 val1 val2
# 1 a 1 4 7
# 2 b 2 5 8
# 3 c 3 6 9
Within each row, I want to calculate the corresponding proportions (ratio) for each value. E.g. for the value in column "val0", I want to calculate row-wise val0 / (val0 + val1 + val2).
Desired output:
id val0 val1 val2
1 a 0.083 0.33 0.583
2 b 0.133 0.33 0.533
3 c 0.167 0.33 0.5
Can anyone tell me what's the best way to do this? Here it's just three columns, but there can be alot of columns.
r dataframe apply
edited Aug 24 '16 at 8:22
Henrik
42.6k994111
asked Apr 16 '13 at 9:02
Rachit AgrawalRachit Agrawal
1,48452148
add a comment |
I have a data frame:
x <- data.frame(id = letters[1:3], val0 = 1:3, val1 = 4:6, val2 = 7:9)
# id val0 val1 val2
# 1 a 1 4 7
# 2 b 2 5 8
# 3 c 3 6 9
Within each row, I want to calculate the corresponding proportions (ratio) for each value. E.g. for the value in column "val0", I want to calculate row-wise val0 / (val0 + val1 + val2).
Desired output:
id val0 val1 val2
1 a 0.083 0.33 0.583
2 b 0.133 0.33 0.533
3 c 0.167 0.33 0.5
Can anyone tell me what's the best way to do this? Here it's just three columns, but there can be alot of columns.
r dataframe apply
edited Aug 24 '16 at 8:22
Henrik
42.6k994111
asked Apr 16 '13 at 9:02
Rachit AgrawalRachit Agrawal
1,48452148
I have a data frame:
x <- data.frame(id = letters[1:3], val0 = 1:3, val1 = 4:6, val2 = 7:9)
# id val0 val1 val2
# 1 a 1 4 7
# 2 b 2 5 8
# 3 c 3 6 9
Within each row, I want to calculate the corresponding proportions (ratio) for each value. E.g. for the value in column "val0", I want to calculate row-wise val0 / (val0 + val1 + val2).
Desired output:
id val0 val1 val2
1 a 0.083 0.33 0.583
2 b 0.133 0.33 0.533
3 c 0.167 0.33 0.5
Can anyone tell me what's the best way to do this? Here it's just three columns, but there can be alot of columns.
r dataframe apply
r dataframe apply
edited Aug 24 '16 at 8:22
Henrik
42.6k994111
asked Apr 16 '13 at 9:02
Rachit AgrawalRachit Agrawal
1,48452148
edited Aug 24 '16 at 8:22
Henrik
42.6k994111
asked Apr 16 '13 at 9:02
Rachit AgrawalRachit Agrawal
1,48452148
edited Aug 24 '16 at 8:22
Henrik
42.6k994111
edited Aug 24 '16 at 8:22
Henrik
42.6k994111
edited Aug 24 '16 at 8:22
Henrik
42.6k994111
42.6k994111
asked Apr 16 '13 at 9:02
Rachit AgrawalRachit Agrawal
1,48452148
asked Apr 16 '13 at 9:02
Rachit AgrawalRachit Agrawal
1,48452148
asked Apr 16 '13 at 9:02
Rachit AgrawalRachit Agrawal
1,48452148
1,48452148
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
And another alternative (though this is mostly a pretty version of sweep)... prop.table:
> cbind(x[1], prop.table(as.matrix(x[-1]), margin = 1))
id val0 val1 val2
1 a 0.08333333 0.3333333 0.5833333
2 b 0.13333333 0.3333333 0.5333333
3 c 0.16666667 0.3333333 0.5000000
From the "description" section of the help file at ?prop.table:
This is really
sweep(x, margin, margin.table(x, margin), "/")for newbies, except that if margin has length zero, then one gets x/sum(x).
So, you can see that underneath, this is really quite similar to @Jilber's solution.
And... it's nice for the R developers to be considerate of us newbies, isn't it? :)
answered Apr 16 '13 at 10:16
A5C1D2H2I1M1N2O1R2T1A5C1D2H2I1M1N2O1R2T1
155k19293388
1
@Jilber, Thanks. Actually, it was inspired by your solution since I always remember the description ofprop.tablestarting with saying that it issweepfor newbies (which I am, perpetually).
– A5C1D2H2I1M1N2O1R2T1
Apr 16 '13 at 11:00
add a comment |
following should do the trick
cbind(id = x[, 1], x[, -1]/rowSums(x[, -1]))
## id val0 val1 val2
## 1 a 0.08333333 0.3333333 0.5833333
## 2 b 0.13333333 0.3333333 0.5333333
## 3 c 0.16666667 0.3333333 0.5000000
answered Apr 16 '13 at 9:06
Chinmay PatilChinmay Patil
13.7k42954
an alternative to usingcbind:x[-1] <- x[-1] / rowSums(x[-1])
– Jaap
Nov 22 '18 at 9:47
add a comment |
Another alternative using sweep
sweep(x[,-1], 1, rowSums(x[,-1]), FUN="/")
val0 val1 val2
1 0.08333333 0.3333333 0.5833333
2 0.13333333 0.3333333 0.5333333
3 0.16666667 0.3333333 0.5000000
answered Apr 16 '13 at 9:19
Jilber UrbinaJilber Urbina
43.6k484114
add a comment |
The function adorn_percentages() from the janitor package does this:
library(janitor)
x %>% adorn_percentages()
id val0 val1 val2
a 0.08333333 0.3333333 0.5833333
b 0.13333333 0.3333333 0.5333333
c 0.16666667 0.3333333 0.5000000
This is equivalent to x %>% adorn_percentages(denominator = "row"), though "row" is the default argument so is not needed in this case. An equivalent call is adorn_percentages(x) if you prefer it without the %>% pipe.
Disclaimer: I created the janitor package, but feel it's appropriate to post this; the function was built to perform exactly this task while making code clearer to read, and the package can be installed from CRAN.
answered Oct 13 '16 at 20:36
Sam FirkeSam Firke
10.5k35569
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
And another alternative (though this is mostly a pretty version of sweep)... prop.table:
> cbind(x[1], prop.table(as.matrix(x[-1]), margin = 1))
id val0 val1 val2
1 a 0.08333333 0.3333333 0.5833333
2 b 0.13333333 0.3333333 0.5333333
3 c 0.16666667 0.3333333 0.5000000
From the "description" section of the help file at ?prop.table:
This is really
sweep(x, margin, margin.table(x, margin), "/")for newbies, except that if margin has length zero, then one gets x/sum(x).
So, you can see that underneath, this is really quite similar to @Jilber's solution.
And... it's nice for the R developers to be considerate of us newbies, isn't it? :)
answered Apr 16 '13 at 10:16
A5C1D2H2I1M1N2O1R2T1A5C1D2H2I1M1N2O1R2T1
155k19293388
1
@Jilber, Thanks. Actually, it was inspired by your solution since I always remember the description ofprop.tablestarting with saying that it issweepfor newbies (which I am, perpetually).
– A5C1D2H2I1M1N2O1R2T1
Apr 16 '13 at 11:00
add a comment |
And another alternative (though this is mostly a pretty version of sweep)... prop.table:
> cbind(x[1], prop.table(as.matrix(x[-1]), margin = 1))
id val0 val1 val2
1 a 0.08333333 0.3333333 0.5833333
2 b 0.13333333 0.3333333 0.5333333
3 c 0.16666667 0.3333333 0.5000000
From the "description" section of the help file at ?prop.table:
This is really
sweep(x, margin, margin.table(x, margin), "/")for newbies, except that if margin has length zero, then one gets x/sum(x).
So, you can see that underneath, this is really quite similar to @Jilber's solution.
And... it's nice for the R developers to be considerate of us newbies, isn't it? :)
answered Apr 16 '13 at 10:16
A5C1D2H2I1M1N2O1R2T1A5C1D2H2I1M1N2O1R2T1
155k19293388
1
@Jilber, Thanks. Actually, it was inspired by your solution since I always remember the description ofprop.tablestarting with saying that it issweepfor newbies (which I am, perpetually).
– A5C1D2H2I1M1N2O1R2T1
Apr 16 '13 at 11:00
add a comment |
And another alternative (though this is mostly a pretty version of sweep)... prop.table:
> cbind(x[1], prop.table(as.matrix(x[-1]), margin = 1))
id val0 val1 val2
1 a 0.08333333 0.3333333 0.5833333
2 b 0.13333333 0.3333333 0.5333333
3 c 0.16666667 0.3333333 0.5000000
From the "description" section of the help file at ?prop.table:
This is really
sweep(x, margin, margin.table(x, margin), "/")for newbies, except that if margin has length zero, then one gets x/sum(x).
So, you can see that underneath, this is really quite similar to @Jilber's solution.
And... it's nice for the R developers to be considerate of us newbies, isn't it? :)
answered Apr 16 '13 at 10:16
A5C1D2H2I1M1N2O1R2T1A5C1D2H2I1M1N2O1R2T1
155k19293388
And another alternative (though this is mostly a pretty version of sweep)... prop.table:
> cbind(x[1], prop.table(as.matrix(x[-1]), margin = 1))
id val0 val1 val2
1 a 0.08333333 0.3333333 0.5833333
2 b 0.13333333 0.3333333 0.5333333
3 c 0.16666667 0.3333333 0.5000000
From the "description" section of the help file at ?prop.table:
This is really
sweep(x, margin, margin.table(x, margin), "/")for newbies, except that if margin has length zero, then one gets x/sum(x).
So, you can see that underneath, this is really quite similar to @Jilber's solution.
And... it's nice for the R developers to be considerate of us newbies, isn't it? :)
answered Apr 16 '13 at 10:16
A5C1D2H2I1M1N2O1R2T1A5C1D2H2I1M1N2O1R2T1
155k19293388
edited Apr 16 '13 at 10:59
answered Apr 16 '13 at 10:16
A5C1D2H2I1M1N2O1R2T1A5C1D2H2I1M1N2O1R2T1
155k19293388
answered Apr 16 '13 at 10:16
A5C1D2H2I1M1N2O1R2T1A5C1D2H2I1M1N2O1R2T1
155k19293388
answered Apr 16 '13 at 10:16
A5C1D2H2I1M1N2O1R2T1A5C1D2H2I1M1N2O1R2T1
155k19293388
155k19293388
1
@Jilber, Thanks. Actually, it was inspired by your solution since I always remember the description ofprop.tablestarting with saying that it issweepfor newbies (which I am, perpetually).
– A5C1D2H2I1M1N2O1R2T1
Apr 16 '13 at 11:00
add a comment |
1
@Jilber, Thanks. Actually, it was inspired by your solution since I always remember the description ofprop.tablestarting with saying that it issweepfor newbies (which I am, perpetually).
– A5C1D2H2I1M1N2O1R2T1
Apr 16 '13 at 11:00
1
1
@Jilber, Thanks. Actually, it was inspired by your solution since I always remember the description of
prop.table starting with saying that it is sweep for newbies (which I am, perpetually).– A5C1D2H2I1M1N2O1R2T1
Apr 16 '13 at 11:00
@Jilber, Thanks. Actually, it was inspired by your solution since I always remember the description of
prop.table starting with saying that it is sweep for newbies (which I am, perpetually).– A5C1D2H2I1M1N2O1R2T1
Apr 16 '13 at 11:00
add a comment |
following should do the trick
cbind(id = x[, 1], x[, -1]/rowSums(x[, -1]))
## id val0 val1 val2
## 1 a 0.08333333 0.3333333 0.5833333
## 2 b 0.13333333 0.3333333 0.5333333
## 3 c 0.16666667 0.3333333 0.5000000
answered Apr 16 '13 at 9:06
Chinmay PatilChinmay Patil
13.7k42954
an alternative to usingcbind:x[-1] <- x[-1] / rowSums(x[-1])
– Jaap
Nov 22 '18 at 9:47
add a comment |
following should do the trick
cbind(id = x[, 1], x[, -1]/rowSums(x[, -1]))
## id val0 val1 val2
## 1 a 0.08333333 0.3333333 0.5833333
## 2 b 0.13333333 0.3333333 0.5333333
## 3 c 0.16666667 0.3333333 0.5000000
answered Apr 16 '13 at 9:06
Chinmay PatilChinmay Patil
13.7k42954
an alternative to usingcbind:x[-1] <- x[-1] / rowSums(x[-1])
– Jaap
Nov 22 '18 at 9:47
add a comment |
following should do the trick
cbind(id = x[, 1], x[, -1]/rowSums(x[, -1]))
## id val0 val1 val2
## 1 a 0.08333333 0.3333333 0.5833333
## 2 b 0.13333333 0.3333333 0.5333333
## 3 c 0.16666667 0.3333333 0.5000000
answered Apr 16 '13 at 9:06
Chinmay PatilChinmay Patil
13.7k42954
following should do the trick
cbind(id = x[, 1], x[, -1]/rowSums(x[, -1]))
## id val0 val1 val2
## 1 a 0.08333333 0.3333333 0.5833333
## 2 b 0.13333333 0.3333333 0.5333333
## 3 c 0.16666667 0.3333333 0.5000000
answered Apr 16 '13 at 9:06
Chinmay PatilChinmay Patil
13.7k42954
answered Apr 16 '13 at 9:06
Chinmay PatilChinmay Patil
13.7k42954
answered Apr 16 '13 at 9:06
Chinmay PatilChinmay Patil
13.7k42954
answered Apr 16 '13 at 9:06
Chinmay PatilChinmay Patil
13.7k42954
13.7k42954
an alternative to usingcbind:x[-1] <- x[-1] / rowSums(x[-1])
– Jaap
Nov 22 '18 at 9:47
add a comment |
an alternative to usingcbind:x[-1] <- x[-1] / rowSums(x[-1])
– Jaap
Nov 22 '18 at 9:47
an alternative to using
cbind: x[-1] <- x[-1] / rowSums(x[-1])– Jaap
Nov 22 '18 at 9:47
an alternative to using
cbind: x[-1] <- x[-1] / rowSums(x[-1])– Jaap
Nov 22 '18 at 9:47
add a comment |
Another alternative using sweep
sweep(x[,-1], 1, rowSums(x[,-1]), FUN="/")
val0 val1 val2
1 0.08333333 0.3333333 0.5833333
2 0.13333333 0.3333333 0.5333333
3 0.16666667 0.3333333 0.5000000
answered Apr 16 '13 at 9:19
Jilber UrbinaJilber Urbina
43.6k484114
add a comment |
Another alternative using sweep
sweep(x[,-1], 1, rowSums(x[,-1]), FUN="/")
val0 val1 val2
1 0.08333333 0.3333333 0.5833333
2 0.13333333 0.3333333 0.5333333
3 0.16666667 0.3333333 0.5000000
answered Apr 16 '13 at 9:19
Jilber UrbinaJilber Urbina
43.6k484114
add a comment |
Another alternative using sweep
sweep(x[,-1], 1, rowSums(x[,-1]), FUN="/")
val0 val1 val2
1 0.08333333 0.3333333 0.5833333
2 0.13333333 0.3333333 0.5333333
3 0.16666667 0.3333333 0.5000000
answered Apr 16 '13 at 9:19
Jilber UrbinaJilber Urbina
43.6k484114
Another alternative using sweep
sweep(x[,-1], 1, rowSums(x[,-1]), FUN="/")
val0 val1 val2
1 0.08333333 0.3333333 0.5833333
2 0.13333333 0.3333333 0.5333333
3 0.16666667 0.3333333 0.5000000
answered Apr 16 '13 at 9:19
Jilber UrbinaJilber Urbina
43.6k484114
answered Apr 16 '13 at 9:19
Jilber UrbinaJilber Urbina
43.6k484114
answered Apr 16 '13 at 9:19
Jilber UrbinaJilber Urbina
43.6k484114
answered Apr 16 '13 at 9:19
Jilber UrbinaJilber Urbina
43.6k484114
43.6k484114
add a comment |
add a comment |
The function adorn_percentages() from the janitor package does this:
library(janitor)
x %>% adorn_percentages()
id val0 val1 val2
a 0.08333333 0.3333333 0.5833333
b 0.13333333 0.3333333 0.5333333
c 0.16666667 0.3333333 0.5000000
This is equivalent to x %>% adorn_percentages(denominator = "row"), though "row" is the default argument so is not needed in this case. An equivalent call is adorn_percentages(x) if you prefer it without the %>% pipe.
Disclaimer: I created the janitor package, but feel it's appropriate to post this; the function was built to perform exactly this task while making code clearer to read, and the package can be installed from CRAN.
answered Oct 13 '16 at 20:36
Sam FirkeSam Firke
10.5k35569
add a comment |
The function adorn_percentages() from the janitor package does this:
library(janitor)
x %>% adorn_percentages()
id val0 val1 val2
a 0.08333333 0.3333333 0.5833333
b 0.13333333 0.3333333 0.5333333
c 0.16666667 0.3333333 0.5000000
This is equivalent to x %>% adorn_percentages(denominator = "row"), though "row" is the default argument so is not needed in this case. An equivalent call is adorn_percentages(x) if you prefer it without the %>% pipe.
Disclaimer: I created the janitor package, but feel it's appropriate to post this; the function was built to perform exactly this task while making code clearer to read, and the package can be installed from CRAN.
answered Oct 13 '16 at 20:36
Sam FirkeSam Firke
10.5k35569
add a comment |
The function adorn_percentages() from the janitor package does this:
library(janitor)
x %>% adorn_percentages()
id val0 val1 val2
a 0.08333333 0.3333333 0.5833333
b 0.13333333 0.3333333 0.5333333
c 0.16666667 0.3333333 0.5000000
This is equivalent to x %>% adorn_percentages(denominator = "row"), though "row" is the default argument so is not needed in this case. An equivalent call is adorn_percentages(x) if you prefer it without the %>% pipe.
Disclaimer: I created the janitor package, but feel it's appropriate to post this; the function was built to perform exactly this task while making code clearer to read, and the package can be installed from CRAN.
answered Oct 13 '16 at 20:36
Sam FirkeSam Firke
10.5k35569
The function adorn_percentages() from the janitor package does this:
library(janitor)
x %>% adorn_percentages()
id val0 val1 val2
a 0.08333333 0.3333333 0.5833333
b 0.13333333 0.3333333 0.5333333
c 0.16666667 0.3333333 0.5000000
This is equivalent to x %>% adorn_percentages(denominator = "row"), though "row" is the default argument so is not needed in this case. An equivalent call is adorn_percentages(x) if you prefer it without the %>% pipe.
Disclaimer: I created the janitor package, but feel it's appropriate to post this; the function was built to perform exactly this task while making code clearer to read, and the package can be installed from CRAN.
answered Oct 13 '16 at 20:36
Sam FirkeSam Firke
10.5k35569
edited Oct 9 '18 at 1:56
answered Oct 13 '16 at 20:36
Sam FirkeSam Firke
10.5k35569
answered Oct 13 '16 at 20:36
Sam FirkeSam Firke
10.5k35569
answered Oct 13 '16 at 20:36
Sam FirkeSam Firke
10.5k35569
10.5k35569
add a comment |
add a comment |
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