问题
I am trying to convert a data frame from wide to long format by gathering specific pairs of columns of which example is shown below:
An example of data frame
df <- data.frame(id=c(1,2,3,4,5), var=c("a","d","g","f","i"),a1=c(3,5,1,2,2), b1=c(2,4,1,2,3), a2=c(8,1,2,5,1), b2=c(1,6,4,7,2), a3=c(7,7,2,3,1), b3=c(1,1,4,9,6))
Initial table:
id var a1 b1 a2 b2 a3 b3
1 1 a 3 2 8 1 7 1
2 2 d 5 4 1 6 7 1
3 3 g 1 1 2 4 2 4
4 4 f 2 2 5 7 3 9
5 5 i 2 3 1 2 1 6
Desired result:
id var a b
1 1 a 3 2
2 1 a 8 1
3 1 a 7 1
4 2 d 5 4
5 2 d 1 6
6 2 d 7 1
7 3 g 1 1
8 3 g 2 4
9 3 g 2 4
10 4 f 2 2
11 4 f 5 7
12 4 f 3 9
13 5 i 2 3
14 5 i 1 2
15 5 i 1 6
Conditions:
- Pair of ai and bi should be gathered: As there are 3 pairs of a and b, "a1 and b1", "a2 and b2" and "a3 and b3", values in those pairs should be moved to a pair of "a and b" by replicating each record in three times
- First and second fields (id of each sample and its common variable) should be kept in each replicated rows
I was thinking that it is possible to make it by gather() in tidyverse, however, as far as I understand, I suppose that gather function may not be suitable for gathering such specific pairs of fields into specific multiple columns (two columns in this case).
It is possible to make it to prepare three data frames separately and binding it into one (example scripts are shown below), however I prefer to make it in one continuous pipe operation in tidyverse not to stop manipulation.
df1 <- df %>% dplyr::select(id,var,a1,b1)
df2 <- df %>% dplyr::select(id,var,a2,b2)
df3 <- df %>% dplyr::select(id,var,a3,b3)
df.fin <- bind_rows(df1,df2,df3)
I would appreciate your elegant suggestons using tidyverse.
=================Additional Questions==================
@Akrun & Camille Thank you for your suggestions and sorry for my late reply. I am now trying to apply your idea into actual data frame but still struggling with another issue.
Followings are column names in actual data frame (sorry, I do not set any values of each columns as it may not be a matter).
colnames(df) <- c("hid","mid","rel","age","gen","mlic","vlic",
"wtaz","staz","ocp","ocpot","emp","empot","expm",
"minc","otaz1","op1","dtime1","atime1","dp1","dtaz1",
"pur1", "repm1","lg1t1","lg2t1","lg3t1","lg4t1","expt1",
"otaz2","op2","dtime2","atime2","dp2","dtaz2","pur2",
"repm2","lg1t2","lg2t2","lg3t2","lg4t2","expt2",
"otaz3","op3","dtime3","atime3","dp3","dtaz3","pur3",
"repm3","lg1t3","lg2t3","lg3t3","lg4t3","expt3",
"otaz4","op4","dtime4","atime4","dp4","dtaz4","pur4",
"repm4","lg1t4","lg2t4","lg3t4","lg4t4","expt4",
"otaz5","op5","dtime5","atime5","dp5","dtaz5","pur5",
"repm5","lg1t5","lg2t5","lg3t5","lg4t5","expt5"
)
Then, I am trying to apply your suggestions as below: In the data frame, columns 1:15 are commons variables and others are repeated variables with 5 repetitions (1 to 5 located at the end of each varible). I could rund following script but still have problem:
#### Convert member table into activity table
## Common variables
hm.com <- names(hm)[c(1:15)]
## Repeating variables
hm.rep <- names(hm)[c(-1:-15)]
hm.rename <- unique(sub("\\d+$","",hm.rep))
## Extract members with trips
hm.trip <- hm %>% filter(otaz!=0) %>% data.frame()
## Convert from member into trip table
test <- split(hm.rep, sub(".*[^1-9$]", "", hm.rep)) %>%
map_df(~ hm.trip %>% dplyr::select(hm.com, .x)) %>%
rename_at(16:28, ~ hm.rename) %>%
arrange(hid,mid,dtime,atime) %>%
data.frame()
The result still have an issue:
I could rename first set of repeated variables, however remaining fields from 2 to 5 are still remaining and records are not appropriately stored in the data frame. I mean that, a set of repeated variables, for instance, from otaz2 to expt2, are stored not in the second row of otaz~expt but stored in its original position (from otaz2 to expt2). I suppose map_df is not working correctly in my case.
========== Problem Solved ========== Above script was containing incorrect manipulation:
Wrong:
map_df(~ hm.trip %>% dplyr::select(hm.com, .x)) %>%
rename_at(16:28, ~ hm.rename)
Correct:
map_df(~ hm.trip %>% dplyr::select(hm.com, .x) %>%
rename_at(16:28, ~ hm.rename))
Thank you, I could go to the next step.
回答1:
This isn't very scaleable, so if you end up needing more than these 3 pairs of columns, go with @akrun's answer. I just wanted to point out that the bind_rows snippet you included could, in fact, be done in one pipe:
library(tidyverse)
bind_rows(
df %>% select(id, var, a = a1, b = b1),
df %>% select(id, var, a = a2, b = b2),
df %>% select(id, var, a = a3, b = b3)
) %>%
arrange(id, var)
#> id var a b
#> 1 1 a 3 2
#> 2 1 a 8 1
#> 3 1 a 7 1
#> 4 2 d 5 4
#> 5 2 d 1 6
#> 6 2 d 7 1
#> 7 3 g 1 1
#> 8 3 g 2 4
#> 9 3 g 2 4
#> 10 4 f 2 2
#> 11 4 f 5 7
#> 12 4 f 3 9
#> 13 5 i 2 3
#> 14 5 i 1 2
#> 15 5 i 1 6
Created on 2018-05-07 by the reprex package (v0.2.0).
If you want something that scales and you like map_* functions (from purrr in the tidyverse), you can abstract the above pipeline:
1:3 %>%
map_df(~select(df, id, var, ends_with(as.character(.))) %>%
setNames(c("id", "var", "a", "b"))) %>%
arrange(id, var)
where 1:3 just represents the numbers of the pairs you have.
回答2:
We could do this with melt from data.table which can take multiple patterns in the measure argument to reshape into 'long' format. In this case we are using column names that start (^) with "a" followed by numbers as one pattern and those start with "b" and followed by numbers as other
library(data.table)
melt(setDT(df), measure = patterns("^a\\d+", "^b\\d+"),
value.name = c("a", "b"))[order(id)][, variable := NULL][]
# id var a b
# 1: 1 a 3 2
# 2: 1 a 8 1
# 3: 1 a 7 1
# 4: 2 d 5 4
# 5: 2 d 1 6
# 6: 2 d 7 1
# 7: 3 g 1 1
# 8: 3 g 2 4
# 9: 3 g 2 4
#10: 4 f 2 2
#11: 4 f 5 7
#12: 4 f 3 9
#13: 5 i 2 3
#14: 5 i 1 2
#15: 5 i 1 6
Or using tidyverse, we gather the columns of interest to 'long' format (but should be cautious when dealing with groups of columns that are having different classes - where melt is more useful), then separate the 'key' column into two, and spread to 'wide' format
library(tidyverse)
df %>%
gather(key, val, a1:b3) %>%
separate(key, into = c("key1", "key2"), sep=1) %>%
spread(key1, val) %>%
select(-key2)
# id var a b
#1 1 a 3 2
#2 1 a 8 1
#3 1 a 7 1
#4 2 d 5 4
#5 2 d 1 6
#6 2 d 7 1
#7 3 g 1 1
#8 3 g 2 4
#9 3 g 2 4
#10 4 f 2 2
#11 4 f 5 7
#12 4 f 3 9
#13 5 i 2 3
#14 5 i 1 2
#15 5 i 1 6
回答3:
a base R solution:
res <- do.call(rbind,lapply(1:3,function(x) setNames(df[c(1:2,2*x+(1:2))],names(df)[1:4])))
res[order(res$id),]
# id var a1 b1
# 1 1 a 3 2
# 6 1 a 8 1
# 11 1 a 7 1
# 2 2 d 5 4
# 7 2 d 1 6
# 12 2 d 7 1
# 3 3 g 1 1
# 8 3 g 2 4
# 13 3 g 2 4
# 4 4 f 2 2
# 9 4 f 5 7
# 14 4 f 3 9
# 5 5 i 2 3
# 10 5 i 1 2
# 15 5 i 1 6
来源：https://stackoverflow.com/questions/50219121/gathering-specific-pairs-of-columns-into-rows-by-dplyr-in-r