进行了一项调查,其中一个问题可以选择多个答案。当选择多个答案时,它们都记录在同一个单元格中。
此外,每个测量员在单元中记录此信息的方式都不同。有时分隔符是连字符 (-),有时是正斜杠 (/)。一些测量员还用数字列出了这些项目。
一个例子是房子里的元素列表(见下图/图片)。我想在每个项目可用时创建列(新列可以有 1/0 或项目名称/NA)(请参见下面的结果示例)。
我可以使用文本到列和查找数组在 Excel 中执行此操作,但是有很多具有同一列的 Excel 工作表,我必须在 R 中执行此操作。抱歉,我不知道如何制作使用 R 代码的示例表,但希望有人能够提供帮助。
数据如下:
House = c("h1","h2","h3","h4","h5","h6","h7","h8","h9","h10","h11")
Items = c("Chair", "Chair- Window/Glass- "," Door- Sofa-", "Chair-
Window/Glass Frame- ", "1. Window/Glass Frame", "Chair- Door- Window-", "Chair- Sofa - Door- Table-", " 4. Table", "Couch (2)", "Window- Table- Chair- Sofa- Door- Couach", "2. Door / Chair")
table1 = as.data.table(House)
table2 = as.data.table(Items)
table = as.data.frame(append(table1, table2))
+-------+------------------------------------------+
| House | Items |
+-------+------------------------------------------+
| 001 | Chair |
| 002 | Chair- Window/Glass- |
| 003 | Door- Sofa- |
| 004 | Chair- Window/Glass Frame- |
| 005 | 1. Window/Glass Frame |
| 006 | Chair- Door- Window- |
| 007 | Chair- Sofa - Door- Table- |
| 008 | 4. Table |
| 009 | Couch (2) |
| 010 | Window- Table- Chair- Sofa- Door- Couach |
| 011 | 2. Door / Chair |
+-------+------------------------------------------+
我的想法是使用所有分隔符(strsplit)进行分割,删除空格(trimws),得到一个唯一列表(unique),然后用我想要的标准(grepl)替换所有变体,最后根据类别。
items <- strsplit(df$Items, "[/.-]")
items <- trimws(items)
items <- df$Items %>%
strsplit("[/.-]") %>%
str_trim(side = "both")
items_list <- unique(items)
这就是我想要得到的: ( window 和玻璃是相同的,椅子/沙发/沙发是相同的,等等 - 所以我只需要创建更大的类别,而不是有几列本质上相同的东西)
+-------+-------+--------+-------+------+
| House | Chair | Window | Table | Door |
+-------+-------+--------+-------+------+
| 001 | Chair | | | |
| 002 | Chair | Window | | |
| 003 | Chair | | | Door |
| 004 | Chair | Window | | |
| 005 | | Window | | |
| 006 | Chair | Window | | Door |
| 007 | Chair | | Table | Door |
| 008 | | | Table | |
| 009 | Chair | | | |
| 010 | Chair | Window | Table | Door |
| 011 | Chair | | | Door |
+-------+-------+--------+-------+------+
最佳答案
您可以在map_df
(或sapply
)中使用str_detect
(或grepl
)来生成以下数据帧:逻辑,将它们强制为整数 0/1,然后将其绑定(bind)到原始数据帧。这种方法绕过了 split /清洁等的麻烦。数据。它只需要您首先为正则表达式创建模式组,即 chair|sofa|couach|couch
、window|glass
:
library(stringr)
library(dplyr)
library(purrr)
# Create regex pattern groups.
patts <- c(chair = "chair|sofa|couach|couch", window = "window|glass",
table = "table", door = "door")
# Detect pattern groups, coerce to 0/1, bind to origional dataframe.
map_df(patts, ~ str_detect(df$Items, regex(., ignore_case = T))) %>%
mutate_all(as.integer) %>%
bind_cols(df, .)
这将返回以下数据帧:
# A tibble: 11 x 6
House Items chair window table door
<dbl> <chr> <int> <int> <int> <int>
1 1 Chair 1 0 0 0
2 2 "Chair- Window/Glass- " 1 1 0 0
3 3 " Door- Sofa-" 1 0 0 1
4 4 "Chair- Window/Glass Frame- " 1 1 0 0
5 5 1. Window/Glass Frame 0 1 0 0
6 6 Chair- Door- Window- 1 1 0 1
7 7 Chair- Sofa - Door- Table- 1 0 1 1
8 8 " 4. Table" 0 0 1 0
9 9 Couch (2) 1 0 0 0
10 10 Window- Table- Chair- Sofa- Door- Couach 1 1 1 1
11 11 2. Door / Chair 1 0 0 1
数据:
df <- tibble(House = c(1,2,3,4,5,6,7,8,9,10,11), Items = c("Chair", "Chair- Window/Glass- "," Door- Sofa-", "Chair- Window/Glass Frame- ", "1. Window/Glass Frame", "Chair- Door- Window-", "Chair- Sofa - Door- Table-", " 4. Table", "Couch (2)", "Window- Table- Chair- Sofa- Door- Couach", "2. Door / Chair"))
关于r - 将不均匀的字符串拆分为 R 中的排序列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56792985/