[英]Gather multiple sets of columns
I have data from an online survey where respondents go through a loop of questions 1-3 times.我有来自在线调查的数据,其中受访者会回答 1-3 次问题循环。 The survey software (Qualtrics) records this data in multiple columns—that is, Q3.2 in the survey will have columns
Q3.2.1.
调查软件 (Qualtrics) 将这些数据记录在多列中——也就是说,调查中的
Q3.2.1.
将包含Q3.2.1.
列Q3.2.1.
, Q3.2.2.
,
Q3.2.2.
, and Q3.2.3.
和
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:我想将所有 QN.N* 列组合成整洁的单独 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,它非常适合组合一组列:
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).正如预期的那样,生成的数据框有 30 行(10 个个体,每个 3 个循环)。 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):然而,收集第二组列不能正常工作——它成功地使两个组合列
Q3.2
和Q3.3
,但最终得到 90 行而不是 30(10 个人的所有组合,Q3.2 的 3 个循环,以及 Q3.3 的 3 个循环;实际数据中每组列的组合将大幅增加):
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?有没有办法像这样使用多次调用
gather()
,在保持正确的行数的同时组合像这样的列的小子集?
This approach seems pretty natural to me:这种方法对我来说似乎很自然:
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.首先收集所有问题列,使用
extract()
分离成question
和loop_number
,然后spread()
问题回到列中。
#> 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
This could be done using reshape
.这可以使用
reshape
来完成。 It is possible with dplyr
though.不过用
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
With new version of tidyr
, we can use pivot_longer
to reshape multiple columns.使用新版本的
tidyr
,我们可以使用pivot_longer
来重塑多个列。 (Using the changed column names from gsub
above) (使用上面
gsub
更改的列名)
library(dplyr)
library(tidyr)
df %>%
pivot_longer(cols = starts_with("Q3"),
names_to = c(".value", "Q3"), names_sep = "_") %>%
select(-Q3)
# A tibble: 30 x 4
# id time Q3.2 Q3.3
# <int> <date> <dbl> <dbl>
# 1 1 2009-01-01 0.974 1.47
# 2 1 2009-01-01 -0.849 -0.513
# 3 1 2009-01-01 0.894 0.0442
# 4 2 2009-01-02 2.04 -0.553
# 5 2 2009-01-02 0.694 0.0972
# 6 2 2009-01-02 -1.11 1.85
# 7 3 2009-01-03 0.413 0.733
# 8 3 2009-01-03 -0.896 -0.271
#9 3 2009-01-03 0.509 -0.0512
#10 4 2009-01-04 1.81 0.668
# … with 20 more rows
NOTE: Values are different because there was no set seed in creating the input dataset注意:值不同是因为在创建输入数据集时没有设置种子
With the recent update to melt.data.table
, we can now melt multiple columns.随着最近对
melt.data.table
更新,我们现在可以融合多个列。 With that, we can do:有了这个,我们可以做到:
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 .您可以从这里获取开发版本。
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.这不是在所有涉及到“tidyr”和“dplyr”,但这里的另一个值得考虑的选择:
merged.stack
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.
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,无法弄清楚如何使用“带捕获组的正则表达式”进行
extract
,以下代码将复制 Hadleys 回答中的extract(...)
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.这里的问题是初始聚集形成一个键列,它实际上是两个键的组合。 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.我选择在评论中的原始解决方案中使用
mutate
将此列拆分为具有等效信息的两列,一个loop_number
列和一个question_number
列。 spread
can then be used to transform the long form data, which are key value pairs (question_number, value)
to wide form data.然后可以使用
spread
将长格式数据(键值对(question_number, value)
为宽格式数据。
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