使用 dplyr 嵌套或分组两个变量，然后对数据执行 Cronbach 的 alpha 函数或其他统计
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【中文标题】使用 dplyr 嵌套或分组两个变量，然后对数据执行 Cronbach 的 alpha 函数或其他统计【英文标题】：Using dplyr to nest or group two variables, then perform the Cronbach's alpha function or other statistics to the data 【发布时间】：2021-11-16 23:55:39 【问题描述】：在心理学中，下面介绍的这种数据集很常见
我想 group 所有年龄（变量 = quest），而不是对所有尺度（com_a4_1:com_a4_6; 和 gm_a4_1:gm_a4_6 等）进行分组，然后对数据应用可靠性函数（psych::alpha） .
我成功创建了这个语法
d %>%
select(quest,contains("_a4_")) %>% #get the data
group_by(quest) %>% #group by all age interval
do(alpha(.)$total)
但是，我无法使用秤的项目“分”嵌套。
据我想象，我必须先对数据进行透视，然后进行分组或嵌套。但是，我在这一点上没有任何成功。我的预期结果类似于下面这张图片。有“两个嵌套的结果”。第一个结果按比例分组（例如：com_a4_1:com_a4_6），第二个结果按年龄分组（quest）
假数据和代码如下
library(psych)
library(tidyverse)
d %>%
select(quest,contains("_a4_")) %>% #get the data
group_by(quest) %>% #group by all age interval
do(alpha(.)$total)
d <-structure(list(quest = c(6, 4, 2, 4, 2, 6, 2, 4, 2, 2, 4, 2,
6, 4, 4, 2, 2, 4, 2, 6, 2, 2, 4, 6, 6, 4, 4, 4, 2, 6, 4, 2, 6,
4, 6, 2, 2, 4, 6, 4, 2), com_a4_1 = c(10, 0, 10, 10, 5, 10, 5,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 5, 10, 10, 0, 10,
10, 10, 10, 10, 5, 10, 10, 10, 10, 10, 10, 10, 10, 5, 10, 10,
10, 10), com_a4_2 = c(10, 10, 5, 10, 10, 5, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 5, 5, 10, 10, 10, 10, 5,
10, 10, 10, 5, 0, 10, 10, 10, 10, 0, 10, 10, 10, 10), com_a4_3 = c(10,
5, 0, 5, 10, 5, 5, 10, 10, 10, 10, 10, 5, 5, 10, 10, 5, 10, 10,
10, 10, 5, 5, 10, 10, 5, 5, 10, 10, 10, 10, 5, 10, 10, 10, 10,
0, 10, 5, 10, 10), com_a4_4 = c(10, 0, 0, 10, 5, 10, 10, 10,
10, 5, 5, 10, 10, 5, 10, 10, 5, 10, 10, 10, 10, 5, 10, 10, 10,
10, 0, 10, 5, 10, 10, 10, 10, 10, 10, 10, 5, 10, 10, 10, 10),
com_a4_5 = c(10, 0, 0, 5, 0, 10, 5, 10, 10, 5, 10, 10, 0,
10, 10, 10, 0, 10, 5, 10, 0, 0, 10, 0, 10, 10, 10, 10, 5,
0, 10, 5, 5, 10, 10, 10, 0, 10, 10, 10, 10), com_a4_6 = c(5,
10, 0, 10, 10, 5, 10, 10, 10, 0, 10, 10, 5, 10, 10, 10, 10,
10, 10, 5, 10, 10, 10, 10, 10, 10, 10, 10, 5, 10, 5, 10,
5, 10, 5, 10, 0, 10, 5, 10, 10), gm_a4_1 = c(10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 5, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10), gm_a4_2 = c(10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 5, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 5, 5, 10, 10, 10, 0, 10, 10,
5, 10, 10, 5, 10, 10, 10, 10), gm_a4_3 = c(10, 10, 10, 10,
10, 5, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 0, 0, 10, 10, 10, 0, 10, 10, 10,
10, 10, 5, 10, 10, 10, 10), gm_a4_4 = c(0, 5, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 5,
10, 10, 10, 10, 10, 0, 0, 10, 10, 10, 0, 10, 5, 5, 5, 10,
10, 10, 10, 10, 10), gm_a4_5 = c(10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 5, 10, 10, 10, 5, 10,
5, 10, 10, 10, 10), gm_a4_6 = c(0, 10, 5, 5, 10, 5, 5, 10,
10, 5, 10, 10, 0, 10, 10, 10, 5, 10, 5, 10, 10, 10, 10, 0,
10, 10, 10, 10, 10, 0, 10, 10, 10, 10, 0, 10, 0, 10, 10,
10, 10), fm_a4_1 = c(10, 5, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 5, 10, 10, 10, 10, 5, 0, 10, 10, 0, 5,
10, 10, 10, 10, 5, 5, 10, 10, 5, 5, 10, 10, 10, 10, 10),
fm_a4_2 = c(10, 10, 10, 10, 0, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 5, 10, 10, 10, 10, 10, 10, 5,
10, 10, 5, 10, 10, 10, 10, 5, 10, 10, 10, 10, 10, 10), fm_a4_3 = c(0,
5, 10, 10, 5, 10, 5, 10, 10, 10, 10, 10, 5, 10, 5, 5, 5,
10, 10, 5, 0, 10, 5, 10, 5, 10, 10, 0, 10, 10, 5, 10, 10,
10, 0, 10, 0, 10, 10, 10, 10), fm_a4_4 = c(10, 5, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 5, 10, 10, 10, 5, 10, 10, 10, 0, 10, 10, 10,
10, 10, 0, 10, 10, 10, 10), fm_a4_5 = c(0, 5, 10, 10, 10,
0, 10, 10, 10, 10, 10, 10, 0, 10, 10, 5, 10, 10, 5, 0, 10,
10, 10, 10, 10, 10, 5, 10, 10, 0, 5, 10, 0, 10, 0, 5, 5,
5, 10, 10, 10), fm_a4_6 = c(10, 5, 5, 0, 0, 5, 10, 10, 10,
0, 10, 10, 5, 10, 10, 10, 0, 10, 0, 10, 10, 0, 10, 10, 5,
0, 0, 10, 10, 10, 0, 10, 10, 5, 5, 10, 0, 0, 10, 10, 5),
cg_a4_1 = c(10, 5, 10, 5, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 0, 10, 10, 10, 10, 5, 0,
10, 10, 10, 10, 5, 10, 10, 10, 10, 5, 5, 10, 10, 10), cg_a4_2 = c(5,
10, 10, 5, 10, 5, 10, 10, 10, 10, 10, 10, 5, 10, 10, 10,
10, 10, 10, 5, 10, 10, 10, 10, 10, 5, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10), cg_a4_3 = c(10,
10, 5, 10, 10, 10, 10, 10, 10, 5, 10, 10, 5, 10, 10, 10,
5, 10, 10, 10, 10, 0, 10, 10, 5, 10, 5, 10, 10, 10, 5, 10,
10, 10, 10, 10, 5, 10, 10, 10, 10), cg_a4_4 = c(10, 10, 0,
5, 5, 5, 10, 10, 10, 5, 10, 10, 0, 5, 10, 10, 5, 10, 10,
10, 10, 0, 5, 10, 10, 5, 0, 0, 10, 10, 0, 10, 0, 10, 10,
5, 0, 5, 5, 10, 10), cg_a4_5 = c(5, 0, 0, 5, 0, 10, 5, 10,
10, 0, 10, 10, 10, 10, 5, 10, 0, 10, 0, 10, 0, 0, 10, 10,
5, 10, 5, 10, 5, 5, 5, 0, 10, 10, 5, 10, 0, 10, 10, 10, 10
), cg_a4_6 = c(0, 0, 5, 10, 10, 10, 10, 10, 0, 10, 5, 10,
10, 10, 5, 10, 10, 10, 10, 10, 5, 10, 10, 10, 10, 5, 5, 10,
5, 10, 0, 10, 10, 5, 5, 10, 5, 10, 10, 10, 10), ps_a4_1 = c(10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 10, 5, 5, 10, 5, 10, 10, 10, 10), ps_a4_2 = c(0, 10,
10, 10, 5, 10, 5, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
10, 5, 10, 5, 10, 10, 10, 5, 10, 10, 10, 5, 0, 10, 10, 10,
5, 0, 10, 5, 10, 10, 10, 10), ps_a4_3 = c(10, 0, 10, 5, 5,
10, 5, 10, 10, 5, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
5, 10, 10, 10, 5, 10, 10, 10, 5, 10, 10, 10, 10, 5, 0, 5,
0, 10, 5, 10, 10), ps_a4_4 = c(10, 10, 10, 10, 5, 10, 5,
10, 10, 0, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 5, 10,
10, 10, 10, 10, 10, 10, 5, 10, 5, 10, 10, 10, 10, 5, 5, 10,
10, 10, 10), ps_a4_5 = c(5, 5, 10, 5, 10, 5, 10, 10, 0, 0,
10, 10, 5, 10, 10, 10, 10, 10, 0, 10, 5, 5, 5, 10, 0, 10,
5, 10, 5, 0, 10, 10, 10, 10, 0, 5, 0, 5, 10, 10, 5), ps_a4_6 = c(5,
5, 0, 5, 0, 10, 0, 10, 5, 5, 10, 10, 5, 10, 10, 10, 0, 10,
5, 10, 5, 0, 5, 10, 5, 10, 5, 0, 5, 10, 0, 0, 10, 5, 0, 5,
0, 10, 10, 10, 10)), row.names = c(NA, -41L), class = "data.frame")
【问题讨论】：
您的预期输出/结果是什么？ 您好！感谢您愿意提供帮助。我编辑了我的问题。 【参考方案1】：我遵循了您的想法，即使用 tidyr 中的pivot_longer() 将比例组按行放置，但将项目保留在列中。 （pivot_longer() 文档中的最后两个示例是我尝试记住如何执行此操作时的首选。）
但是，这取决于您为每个量表拥有相同数量的项目；我不确定它对于每个比例的不同项目如何支持。
一旦事物的形式较长，请在 quest 上使用 nest_by() 和 scales 变量，后跟 mutate() 嵌套并计算每一行的 alpha。
我没有在此处粘贴所有警告和消息，但有很多内容。如果您不再需要它，也可以删除末尾的 data 列。
library(psych)
library(dplyr)
library(tidyr)
d %>%
pivot_longer(cols = -quest,
names_to = c("scale", ".value"),
names_pattern = "(\\w+_\\w+_)(.)") %>%
nest_by(quest, scale) %>%
mutate(alpha(data)$total)
#> # A tibble: 15 x 12
#> # Rowwise: quest, name
#> quest name data raw_alpha std.alpha `G6(smc)` average_r `S/N` ase
#> <dbl> <chr> <list<t> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 2 cg_a4_ [16 x 6] 0.619 0.594 0.728 0.226 1.46 0.141
#> 2 2 com_a~ [16 x 6] 0.810 0.808 0.881 0.412 4.20 0.0719
#> 3 2 fm_a4_ [16 x 6] 0.400 0.421 0.546 0.108 0.728 0.221
#> 4 2 gm_a4_ [16 x 6] 0.842 0.952 0.745 0.831 19.7 0.0592
#> 5 2 ps_a4_ [16 x 6] 0.684 0.753 0.870 0.337 3.05 0.123
#> 6 4 cg_a4_ [15 x 6] 0.677 0.696 0.807 0.276 2.29 0.126
#> 7 4 com_a~ [15 x 6] 0.673 0.613 0.842 0.209 1.58 0.110
#> 8 4 fm_a4_ [15 x 6] 0.669 0.714 0.811 0.294 2.50 0.124
#> 9 4 gm_a4_ [15 x 6] 0.811 0.759 0.873 0.386 3.15 0.0389
#> 10 4 ps_a4_ [15 x 6] 0.533 0.551 0.605 0.170 1.23 0.161
#> 11 6 cg_a4_ [10 x 6] -0.168 -0.00601 0.550 -0.00120 -0.00597 0.621
#> 12 6 com_a~ [10 x 6] -0.184 0.228 0.486 0.0686 0.295 0.644
#> 13 6 fm_a4_ [10 x 6] 0.508 0.542 0.727 0.191 1.18 0.248
#> 14 6 gm_a4_ [10 x 6] -0.075 -0.492 -0.0806 -0.0582 -0.330 0.398
#> 15 6 ps_a4_ [10 x 6] 0.844 0.879 0.903 0.592 7.26 0.0710
#> # ... with 3 more variables: mean <dbl>, sd <dbl>, median_r <dbl>
由reprex package (v2.0.0) 于 2021 年 9 月 23 日创建
【讨论】：
哇！！我印象深刻！您的代码非常干净和有用。太感谢了！效果很好！！【参考方案2】：感觉分组/透视是试图过度设计解决方案。一种方法是编写一个函数，允许您设置位于contains() 中的值。
library(psych)
library(tidyverse)
apply_alpha <- function(data, nest_contains)
data %>%
select(quest, contains(nest_contains)) %>%
group_by(quest) %>%
do(alpha(.)$total)
apply_alpha(d, 'com_')
apply_alpha(d, 'gm_')
apply_alpha(d, 'fm_')
重要的是要注意，通过这种方法，我会收到大量我不熟悉的警告消息。它们来自alpha() 函数的使用。
【讨论】：
【参考方案3】：您可以这样做：让我知道以解释这是否是您正在寻找的内容：
library(tidyverse)
library(psych)
reg_fm_a4_ <- "^fm_a4_.*"
reg_com_a4_ <- "^com_a4_.*"
reg_gm_a4_ <- "^gm_a4_.*"
reg_cg_a4_ <- "^cg_a4_.*"
reg_ps_a4_ <- "^ps_a4_.*"
regs <- c(reg_fm_a4_, reg_com_a4_, reg_gm_a4_, reg_cg_a4_, reg_ps_a4_) %>%
set_names(c("fm_a4_", "com_a4_", "gm_a4_", "cg_a4_",
"ps_a4_"))
cronbachs_alpha <-
map_df(regs, ~
d %>%
select(dplyr::matches(.x)) %>%
psych::alpha(check.keys = TRUE) %>% .$total %>%
tibble::rownames_to_column()
,.id = "scale"
)
scale rowname raw_alpha std.alpha G6(smc) average_r S/N ase mean sd median_r
1 fm_a4_ 0.4655172 0.4841889 0.5081686 0.1352840 0.9386944 0.12722395 8.008130 1.716728 0.16047102
2 com_a4_ 0.7246145 0.7294824 0.7367755 0.3100766 2.6966174 0.06440378 8.130081 2.056329 0.32419199
3 gm_a4_ 0.6285083 0.6818823 0.7360522 0.2632152 2.1434909 0.08701602 7.337398 1.516341 0.09958706
4 cg_a4_ 0.5260735 0.5134628 0.5966499 0.1495805 1.0553414 0.10814655 6.524390 1.737080 0.12196703
5 ps_a4_ 0.7173328 0.7486200 0.7597498 0.3317028 2.9780417 0.06479382 7.906504 1.990620 0.36281243
【讨论】：
快到了！！分组变量（问卷quest）缺失【参考方案4】：
您可以进行重塑，然后使用嵌套数据。当然，如果您不想在结果中保留嵌套数据，您可以取消选择 data 列。
此解决方案的优点（如果您这样看的话）是您 a) 不需要创建额外的对象，也不需要创建特定的函数。
d %>%
mutate(id = 1:n()) %>%
pivot_longer(cols = c(-id, -quest)) %>%
separate(col = name,
into = c("scale", "item"),
sep = "_",
extra = "merge") %>%
pivot_wider(names_from = item) %>%
select(-id) %>%
group_by(quest, scale) %>%
nest() %>%
mutate(alpha_results = map(data, ~alpha(.)$total)) %>%
unnest_wider(alpha_results) %>%
arrange(scale, quest)
给出：
# Groups: quest, scale [15]
quest scale data raw_alpha std.alpha `G6(smc)` average_r `S/N` ase mean sd median_r
<dbl> <chr> <list> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 2 cg <tibble [16 x 6]> 0.619 0.594 0.728 0.226 1.46 0.141 7.60 1.90 0.157
2 4 cg <tibble [15 x 6]> 0.677 0.696 0.807 0.276 2.29 0.126 8 1.88 0.368
3 6 cg <tibble [10 x 6]> -0.168 -0.00601 0.550 -0.00120 -0.00597 0.621 8.42 1.07 0.102
4 2 com <tibble [16 x 6]> 0.810 0.808 0.881 0.412 4.20 0.0719 7.24 2.63 0.457
5 4 com <tibble [15 x 6]> 0.673 0.613 0.842 0.209 1.58 0.110 8.83 1.60 0.201
6 6 com <tibble [10 x 6]> -0.184 0.228 0.486 0.0686 0.295 0.644 8.5 0.946 0.0970
7 2 fm <tibble [16 x 6]> 0.4 0.421 0.546 0.108 0.728 0.221 8.12 1.62 0.0953
8 4 fm <tibble [15 x 6]> 0.669 0.714 0.811 0.294 2.50 0.124 8.28 1.82 0.366
9 6 fm <tibble [10 x 6]> 0.508 0.542 0.727 0.191 1.18 0.248 7.42 1.73 0.279
10 2 gm <tibble [16 x 6]> 0.842 0.952 0.745 0.831 19.7 0.0592 9.48 1.05 0.831
11 4 gm <tibble [15 x 6]> 0.811 0.759 0.873 0.386 3.15 0.0389 8.83 1.94 0.511
12 6 gm <tibble [10 x 6]> -0.075 -0.492 -0.0806 -0.0582 -0.330 0.398 8.33 1.11 -0.111
13 2 ps <tibble [16 x 6]> 0.684 0.753 0.870 0.337 3.05 0.123 7.08 1.97 0.316
14 4 ps <tibble [15 x 6]> 0.533 0.551 0.605 0.170 1.23 0.161 8.83 1.29 0.150
15 6 ps <tibble [10 x 6]> 0.844 0.879 0.903 0.592 7.26 0.0710 7.83 2.43 0.604
【讨论】：
这正是我想要做的！我按照您对“单独”论点的答复。然后我正在努力寻找如何前进！谢谢！以上是关于使用 dplyr 嵌套或分组两个变量，然后对数据执行 Cronbach 的 alpha 函数或其他统计的主要内容，如果未能解决你的问题，请参考以下文章
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