R:如何将字符串拆分为多个值并将生成的碎片作为列映射到数据集?

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问题描述

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如上图所示，我有一个流派"列，其中包含相应电影所属的流派列表.共有19种独特的流派.我想知道我是否可以处理这些数据，将 19 列附加到数据集，每列对应于每个流派标识符，并将相应的单元格标记为 0 或 1，指示电影从属到每个流派列.

As shown in the above pic, I've a column, genres, with a list of genres the corresponding movie belongs to. There are in total 19 unique genres. I'd like to know if I can manipulate this data into appending 19 columns to the data set each corresponding to each of the genres identifiers and label the corresponding cells as 0 or 1 indicating the movies affiliation to the each genre columns.

它应该类似于下图.

推荐答案

我们可以在拆分流派"列后进行此操作

We can do this after splitting the 'genres' column

library(qdapTools)
d1 <- mtabulate(strsplit(as.character(df1$genres),","))
row.names(d1) <- sub("\\s*\\(.*", "", df1$title)

<小时>

或者另一种选择是创建一个列名为流派"的矩阵，然后对拆分的字符串进行比较

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Or another option is to create a matrix with column names as 'genres' and then do a comparison on the splitted string

m1 <- matrix(0, dimnames = list(sub("\\s*\\(.*", "", df1$title), 
      c("Adventure", "Animation", "Children",
   "Comedy", "Fantasy", "Romance", "Action", "Crime", "Thriller")), ncol=9, nrow = nrow(df1))
m1 + (t(sapply(strsplit(as.character(df1$genres), ","), function(x) colnames(m1) %in% x)))
#         Adventure Animation Children Comedy Fantasy Romance Action Crime Thriller
#Toy Story         1         1        1      1       1       0      0     0        0
#Jumanji           1         0        1      0       1       0      0     0        0
#Heat              0         0        0      0       0       0      1     1        1

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