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这个问题已经有了答案:
- Reshaping multiple sets of measurement columns (wide format) into single columns (long format) 6 answers
- 将多组测量列(宽格式)重新形成单列(长格式)6个答案
I have data from an online survey where respondents go through a loop of questions 1-3 times. The survey software (Qualtrics) records this data in multiple columns—that is, Q3.2 in the survey will have columns Q3.2.1., Q3.2.2., and Q3.2.3.:
我有一项在线调查的数据，调查对象会在回答1-3次问题时进行循环。调查软件(Qualtrics)在多个专栏中记录了这一数据——也就是说，调查中的Q3.2将有专栏Q3.2.1。,Q3.2.2。,Q3.2.3。:
df <- data.frame(
id = 1:10,
time = as.Date('2009-01-01') + 0:9,
Q3.2.1. = rnorm(10, 0, 1),
Q3.2.2. = rnorm(10, 0, 1),
Q3.2.3. = rnorm(10, 0, 1),
Q3.3.1. = rnorm(10, 0, 1),
Q3.3.2. = rnorm(10, 0, 1),
Q3.3.3. = rnorm(10, 0, 1)
)
# Sample data
id time Q3.2.1. Q3.2.2. Q3.2.3. Q3.3.1. Q3.3.2. Q3.3.3.
1 1 2009-01-01 -0.2059165 -0.29177677 -0.7107192 1.52718069 -0.4484351 -1.21550600
2 2 2009-01-02 -0.1981136 -1.19813815 1.1750200 -0.40380049 -1.8376094 1.03588482
3 3 2009-01-03 0.3514795 -0.27425539 1.1171712 -1.02641801 -2.0646661 -0.35353058
...
I want to combine all the QN.N* columns into tidy individual QN.N columns, ultimately ending up with something like this:
我想把所有的QN组合起来。N*列组成整洁的独立QN。N列，最终得到这样的结果
id time loop_number Q3.2 Q3.3
1 1 2009-01-01 1 -0.20591649 1.52718069
2 2 2009-01-02 1 -0.19811357 -0.40380049
3 3 2009-01-03 1 0.35147949 -1.02641801
...
11 1 2009-01-01 2 -0.29177677 -0.4484351
12 2 2009-01-02 2 -1.19813815 -1.8376094
13 3 2009-01-03 2 -0.27425539 -2.0646661
...
21 1 2009-01-01 3 -0.71071921 -1.21550600
22 2 2009-01-02 3 1.17501999 1.03588482
23 3 2009-01-03 3 1.11717121 -0.35353058
...
The tidyr library has the gather() function, which works great for combining one set of columns:
tidyr库具有gather()函数，它对组合一组列非常有用:
library(dplyr)
library(tidyr)
library(stringr)
df %>% gather(loop_number, Q3.2, starts_with("Q3.2")) %>%
mutate(loop_number = str_sub(loop_number,-2,-2)) %>%
select(id, time, loop_number, Q3.2)
id time loop_number Q3.2
1 1 2009-01-01 1 -0.20591649
2 2 2009-01-02 1 -0.19811357
3 3 2009-01-03 1 0.35147949
...
29 9 2009-01-09 3 -0.58581232
30 10 2009-01-10 3 -2.33393981
The resultant data frame has 30 rows, as expected (10 individuals, 3 loops each). However, gathering a second set of columns does not work correctly—it successfully makes the two combined columns Q3.2 and Q3.3, but ends up with 90 rows instead of 30 (all combinations of 10 individuals, 3 loops of Q3.2, and 3 loops of Q3.3; the combinations will increase substantially for each group of columns in the actual data):
结果数据帧有30行，如预期的(10个人，3个循环)。但是，收集第二组列并不正确——它成功地使两个组合列Q3.2和Q3.3组合在一起，但是最后得到90行而不是30行(10个个体的所有组合，Q3.2的3个循环，Q3.3的3个循环);实际数据中的每一组列的组合将大量增加):
df %>% gather(loop_number, Q3.2, starts_with("Q3.2")) %>%
gather(loop_number, Q3.3, starts_with("Q3.3")) %>%
mutate(loop_number = str_sub(loop_number,-2,-2))
id time loop_number Q3.2 Q3.3
1 1 2009-01-01 1 -0.20591649 1.52718069
2 2 2009-01-02 1 -0.19811357 -0.40380049
3 3 2009-01-03 1 0.35147949 -1.02641801
...
89 9 2009-01-09 3 -0.58581232 -0.13187024
90 10 2009-01-10 3 -2.33393981 -0.48502131
Is there a way to use multiple calls to gather() like this, combining small subsets of columns like this while maintaining the correct number of rows?
是否有一种方法可以像这样使用多个调用collect()，同时组合像这样的小列子集，同时保持正确的行数?
5 个解决方案
#1
97
This approach seems pretty natural to me:
这种方法在我看来很自然:
df %>%
gather(key, value, -id, -time) %>%
extract(key, c("question", "loop_number"), "(Q.\\..)\\.(.)") %>%
spread(question, value)
First gather all question columns, use extract() to separate into question and loop_number, then spread() question back into the columns.
首先收集所有的问题列，使用extract()将问题和loop_number分开，然后将()问题传播回列中。
#> id time loop_number Q3.2 Q3.3
#> 1 1 2009-01-01 1 0.142259203 -0.35842736
#> 2 1 2009-01-01 2 0.061034802 0.79354061
#> 3 1 2009-01-01 3 -0.525686204 -0.67456611
#> 4 2 2009-01-02 1 -1.044461185 -1.19662936
#> 5 2 2009-01-02 2 0.393808163 0.42384717
#2
23
This could be done using reshape. It is possible with dplyr though.
这可以用整形来完成。尽管dplyr是可能的。
colnames(df) <- gsub("\\.(.{2})$", "_\\1", colnames(df))
colnames(df)[2] <- "Date"
res <- reshape(df, idvar=c("id", "Date"), varying=3:8, direction="long", sep="_")
row.names(res) <- 1:nrow(res)
head(res)
# id Date time Q3.2 Q3.3
#1 1 2009-01-01 1 1.3709584 0.4554501
#2 2 2009-01-02 1 -0.5646982 0.7048373
#3 3 2009-01-03 1 0.3631284 1.0351035
#4 4 2009-01-04 1 0.6328626 -0.6089264
#5 5 2009-01-05 1 0.4042683 0.5049551
#6 6 2009-01-06 1 -0.1061245 -1.7170087
Or using dplyr
或者使用dplyr
library(tidyr)
library(dplyr)
colnames(df) <- gsub("\\.(.{2})$", "_\\1", colnames(df))
df %>%
gather(loop_number, "Q3", starts_with("Q3")) %>%
separate(loop_number,c("L1", "L2"), sep="_") %>%
spread(L1, Q3) %>%
select(-L2) %>%
head()
# id time Q3.2 Q3.3
#1 1 2009-01-01 1.3709584 0.4554501
#2 1 2009-01-01 1.3048697 0.2059986
#3 1 2009-01-01 -0.3066386 0.3219253
#4 2 2009-01-02 -0.5646982 0.7048373
#5 2 2009-01-02 2.2866454 -0.3610573
#6 2 2009-01-02 -1.7813084 -0.7838389
#3
16
With the recent update to melt.data.table, we can now melt multiple columns. With that, we can do:
最近对melt.data的更新。表，我们现在可以熔化多列。有了它，我们可以:
require(data.table) ## 1.9.5
melt(setDT(df), id=1:2, measure=patterns("^Q3.2", "^Q3.3"),
value.name=c("Q3.2", "Q3.3"), variable.name="loop_number")
# id time loop_number Q3.2 Q3.3
# 1: 1 2009-01-01 1 -0.433978480 0.41227209
# 2: 2 2009-01-02 1 -0.567995351 0.30701144
# 3: 3 2009-01-03 1 -0.092041353 -0.96024077
# 4: 4 2009-01-04 1 1.137433487 0.60603396
# 5: 5 2009-01-05 1 -1.071498263 -0.01655584
# 6: 6 2009-01-06 1 -0.048376809 0.55889996
# 7: 7 2009-01-07 1 -0.007312176 0.69872938
You can get the development version from here.
您可以从这里获得开发版本。
#4
11
It's not at all related to "tidyr" and "dplyr", but here's another option to consider: merged.stack from my "splitstackshape" package, V1.4.0 and above.
它与“tidyr”和“dplyr”没有任何关系，但是这里有另一个可考虑的选项:合并。堆叠从我的“splitstackshape”包，V1.4.0及以上。
library(splitstackshape)
merged.stack(df, id.vars = c("id", "time"),
var.stubs = c("Q3.2.", "Q3.3."),
sep = "var.stubs")
# id time .time_1 Q3.2. Q3.3.
# 1: 1 2009-01-01 1. -0.62645381 1.35867955
# 2: 1 2009-01-01 2. 1.51178117 -0.16452360
# 3: 1 2009-01-01 3. 0.91897737 0.39810588
# 4: 2 2009-01-02 1. 0.18364332 -0.10278773
# 5: 2 2009-01-02 2. 0.38984324 -0.25336168
# 6: 2 2009-01-02 3. 0.78213630 -0.61202639
# 7: 3 2009-01-03 1. -0.83562861 0.38767161
# <<:::SNIP:::>>
# 24: 8 2009-01-08 3. -1.47075238 -1.04413463
# 25: 9 2009-01-09 1. 0.57578135 1.10002537
# 26: 9 2009-01-09 2. 0.82122120 -0.11234621
# 27: 9 2009-01-09 3. -0.47815006 0.56971963
# 28: 10 2009-01-10 1. -0.30538839 0.76317575
# 29: 10 2009-01-10 2. 0.59390132 0.88110773
# 30: 10 2009-01-10 3. 0.41794156 -0.13505460
# id time .time_1 Q3.2. Q3.3.
#5
6
In case you are like me, and cannot work out how to use "regular expression with capturing groups" for extract, the following code replicates the extract(...) line in Hadleys' answer:
如果您像我一样，无法找到如何使用“带捕获组的正则表达式”进行提取，下面的代码将复制hadley的答案中的提取(…)行:
df %>%
gather(question_number, value, starts_with("Q3.")) %>%
mutate(loop_number = str_sub(question_number,-2,-2), question_number = str_sub(question_number,1,4)) %>%
select(id, time, loop_number, question_number, value) %>%
spread(key = question_number, value = value)
The problem here is that the initial gather forms a key column that is actually a combination of two keys. I chose to use mutate in my original solution in the comments to split this column into two columns with equivalent info, a loop_number column and a question_number column. spread can then be used to transform the long form data, which are key value pairs (question_number, value) to wide form data.
这里的问题是，初始聚集形成一个键列，实际上是两个键的组合。我选择在注释中的原始解决方案中使用mutate，将这个列分为两个列，具有相同的信息，一个loop_number列和一个question_number列。然后可以使用spread将长表单数据转换为宽表单数据，长表单数据是键值对(question_number, value)。
#1
97
This approach seems pretty natural to me:
这种方法在我看来很自然:
df %>%
gather(key, value, -id, -time) %>%
extract(key, c("question", "loop_number"), "(Q.\\..)\\.(.)") %>%
spread(question, value)
First gather all question columns, use extract() to separate into question and loop_number, then spread() question back into the columns.
首先收集所有的问题列，使用extract()将问题和loop_number分开，然后将()问题传播回列中。
#> id time loop_number Q3.2 Q3.3
#> 1 1 2009-01-01 1 0.142259203 -0.35842736
#> 2 1 2009-01-01 2 0.061034802 0.79354061
#> 3 1 2009-01-01 3 -0.525686204 -0.67456611
#> 4 2 2009-01-02 1 -1.044461185 -1.19662936
#> 5 2 2009-01-02 2 0.393808163 0.42384717
#2
23
This could be done using reshape. It is possible with dplyr though.
这可以用整形来完成。尽管dplyr是可能的。
colnames(df) <- gsub("\\.(.{2})$", "_\\1", colnames(df))
colnames(df)[2] <- "Date"
res <- reshape(df, idvar=c("id", "Date"), varying=3:8, direction="long", sep="_")
row.names(res) <- 1:nrow(res)
head(res)
# id Date time Q3.2 Q3.3
#1 1 2009-01-01 1 1.3709584 0.4554501
#2 2 2009-01-02 1 -0.5646982 0.7048373
#3 3 2009-01-03 1 0.3631284 1.0351035
#4 4 2009-01-04 1 0.6328626 -0.6089264
#5 5 2009-01-05 1 0.4042683 0.5049551
#6 6 2009-01-06 1 -0.1061245 -1.7170087
Or using dplyr
或者使用dplyr
library(tidyr)
library(dplyr)
colnames(df) <- gsub("\\.(.{2})$", "_\\1", colnames(df))
df %>%
gather(loop_number, "Q3", starts_with("Q3")) %>%
separate(loop_number,c("L1", "L2"), sep="_") %>%
spread(L1, Q3) %>%
select(-L2) %>%
head()
# id time Q3.2 Q3.3
#1 1 2009-01-01 1.3709584 0.4554501
#2 1 2009-01-01 1.3048697 0.2059986
#3 1 2009-01-01 -0.3066386 0.3219253
#4 2 2009-01-02 -0.5646982 0.7048373
#5 2 2009-01-02 2.2866454 -0.3610573
#6 2 2009-01-02 -1.7813084 -0.7838389
#3
16
With the recent update to melt.data.table, we can now melt multiple columns. With that, we can do:
最近对melt.data的更新。表，我们现在可以熔化多列。有了它，我们可以:
require(data.table) ## 1.9.5
melt(setDT(df), id=1:2, measure=patterns("^Q3.2", "^Q3.3"),
value.name=c("Q3.2", "Q3.3"), variable.name="loop_number")
# id time loop_number Q3.2 Q3.3
# 1: 1 2009-01-01 1 -0.433978480 0.41227209
# 2: 2 2009-01-02 1 -0.567995351 0.30701144
# 3: 3 2009-01-03 1 -0.092041353 -0.96024077
# 4: 4 2009-01-04 1 1.137433487 0.60603396
# 5: 5 2009-01-05 1 -1.071498263 -0.01655584
# 6: 6 2009-01-06 1 -0.048376809 0.55889996
# 7: 7 2009-01-07 1 -0.007312176 0.69872938
You can get the development version from here.
您可以从这里获得开发版本。
#4
11
It's not at all related to "tidyr" and "dplyr", but here's another option to consider: merged.stack from my "splitstackshape" package, V1.4.0 and above.
它与“tidyr”和“dplyr”没有任何关系，但是这里有另一个可考虑的选项:合并。堆叠从我的“splitstackshape”包，V1.4.0及以上。
library(splitstackshape)
merged.stack(df, id.vars = c("id", "time"),
var.stubs = c("Q3.2.", "Q3.3."),
sep = "var.stubs")
# id time .time_1 Q3.2. Q3.3.
# 1: 1 2009-01-01 1. -0.62645381 1.35867955
# 2: 1 2009-01-01 2. 1.51178117 -0.16452360
# 3: 1 2009-01-01 3. 0.91897737 0.39810588
# 4: 2 2009-01-02 1. 0.18364332 -0.10278773
# 5: 2 2009-01-02 2. 0.38984324 -0.25336168
# 6: 2 2009-01-02 3. 0.78213630 -0.61202639
# 7: 3 2009-01-03 1. -0.83562861 0.38767161
# <<:::SNIP:::>>
# 24: 8 2009-01-08 3. -1.47075238 -1.04413463
# 25: 9 2009-01-09 1. 0.57578135 1.10002537
# 26: 9 2009-01-09 2. 0.82122120 -0.11234621
# 27: 9 2009-01-09 3. -0.47815006 0.56971963
# 28: 10 2009-01-10 1. -0.30538839 0.76317575
# 29: 10 2009-01-10 2. 0.59390132 0.88110773
# 30: 10 2009-01-10 3. 0.41794156 -0.13505460
# id time .time_1 Q3.2. Q3.3.
#5
6
In case you are like me, and cannot work out how to use "regular expression with capturing groups" for extract, the following code replicates the extract(...) line in Hadleys' answer:
如果您像我一样，无法找到如何使用“带捕获组的正则表达式”进行提取，下面的代码将复制hadley的答案中的提取(…)行:
df %>%
gather(question_number, value, starts_with("Q3.")) %>%
mutate(loop_number = str_sub(question_number,-2,-2), question_number = str_sub(question_number,1,4)) %>%
select(id, time, loop_number, question_number, value) %>%
spread(key = question_number, value = value)
The problem here is that the initial gather forms a key column that is actually a combination of two keys. I chose to use mutate in my original solution in the comments to split this column into two columns with equivalent info, a loop_number column and a question_number column. spread can then be used to transform the long form data, which are key value pairs (question_number, value) to wide form data.
这里的问题是，初始聚集形成一个键列，实际上是两个键的组合。我选择在注释中的原始解决方案中使用mutate，将这个列分为两个列，具有相同的信息，一个loop_number列和一个question_number列。然后可以使用spread将长表单数据转换为宽表单数据，长表单数据是键值对(question_number, value)。