通过列选择将数据帧拆分为多个数据帧

问题描述

这些是我的数据框:

# data
set.seed(1234321)

# Original data frame (i.e. a questionnaire survey data)
answer <- c("Yes", "No")
likert_scale <- c("strongly disagree", "disagree", "undecided", "agree", "strongly agree")
d1 <- c(rnorm(10)*10)
d2 <- sample(x = c(letters), size = 10, replace = TRUE)
d3 <- sample(x = likert_scale, size = 10, replace = TRUE)
d4 <- sample(x = likert_scale, size = 10, replace = TRUE)
d5 <- sample(x = likert_scale, size = 10, replace = TRUE)
d6 <- sample(x = answer, size = 10, replace = TRUE)
d7 <- sample(x = answer, size = 10, replace = TRUE)
original_df <- data.frame(d1, d2, d3, d4, d5, d6, d7)

# Questionnaire codebook data frame
quest_section <- c("generic", "likert scale", "specific approval")
starting_column <- c(1, 3, 6)
ending_column <- c(2, 5, 7)
df_codebook <- data.frame(quest_section, min_column, max_column)

我想根据 df_codebook 中的 quest_section 变量，使用 starting_column 和 将原始数据帧拆分为不同的数据帧>ending_column 作为 indeces 来选择 original_df 中的列.

I would like to split the orginal dataframe in different ones on the basis of quest_section variable in the df_codebook, using starting_column and ending_column as indeces to select columns in the original_df.

这是我尝试创建一个函数以拆分original_df:

This is what I tried creating a function in order to split the original_df:

# splitting dataframe function
split_df <- function(my_df, my_codebook) {
        df_names <- df_codebook[,1] %>%
                map(set_names)
        for (i in 1:length(df_codebook[,1])) {
                df_names$`[i]` <- original_df %>%
                        dplyr::select(df_codebook[[2]][i]:df_codebook[[3]][i])
        }
}

# apply function to two dataframes
my_df_list <- split_df(my_df = original_df, my_codebook = df_codebook)

结果是一个 NULL 对象而不是以下列表:

and the result was a NULL object instead of the following list:

> my_df_list
$generic
           d1 d2
1   12.369081  z
2   15.616230  x
3   18.396185  f
4    3.173245  q
5   10.715115  j
6  -11.459955  p
7    2.488894  j
8    1.158625  n
9   26.200816  a
10  12.624048  b

$`likert scale`
                  d3                d4                d5
1           disagree    strongly agree    strongly agree
2          undecided         undecided strongly disagree
3     strongly agree         undecided strongly disagree
4              agree         undecided         undecided
5  strongly disagree             agree         undecided
6           disagree strongly disagree         undecided
7           disagree             agree          disagree
8           disagree strongly disagree         undecided
9          undecided strongly disagree          disagree
10 strongly disagree          disagree    strongly agree

$`specific approval`
    d6  d7
1   No  No
2   No  No
3  Yes  No
4  Yes Yes
5  Yes Yes
6  Yes Yes
7  Yes  No
8   No Yes
9   No  No
10  No Yes

我对任何类型的解决方案都感兴趣:使用 tidyverse 和 purrr 方法，或功能性方法.

I am interested in any kind of solution: using tidyverse and purrr approach, or functional one.

推荐答案

您可以使用 Map 在每个 starting_column 之间创建一个序列:ending_column 并使用该序列从 original_df 中提取相关列.我们可以使用 setNames 为列表分配名称.

You can use Map to create a sequence between each of starting_column: ending_column and use that sequence to extract the relevant columns from original_df. We can use setNames to assign names to the list.

setNames(Map(function(x, y) original_df[, x:y],
             df_codebook$starting_column, df_codebook$ending_column),
         df_codebook$quest_section)

返回

#$generic
#           d1 d2
#1   12.369081  z
#2   15.616230  x
#3   18.396185  f
#4    3.173245  q
#5   10.715115  j
#6  -11.459955  p
#7    2.488894  j
#8    1.158625  n
#9   26.200816  a
#10  12.624048  b

#$`likert scale`
#                  d3                d4                d5
#1           disagree    strongly agree    strongly agree
#2          undecided         undecided strongly disagree
#3     strongly agree         undecided strongly disagree
#4              agree         undecided         undecided
#5  strongly disagree             agree         undecided
#6           disagree strongly disagree         undecided
#7           disagree             agree          disagree
#8           disagree strongly disagree         undecided
#9          undecided strongly disagree          disagree
#10 strongly disagree          disagree    strongly agree

#$`specific approval`
#    d6  d7
#1   No  No
#2   No  No
#3  Yes  No
#4  Yes Yes
#5  Yes Yes
#6  Yes Yes
#7  Yes  No
#8   No Yes
#9   No  No
#10  No Yes

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