简体   繁体   English

R:将分隔的字符串转换为变量

[英]R: Convert delimited string into variables

I have a data frame with a column containing a space-delimited list of character codes: 我有一个数据框,其中一列包含以空格分隔的字符代码列表:

"Ab B C"
""
"X C"
"N Ab F S"
:

I want to convert this into multiple columns, one for each distinct value, indicating (with 1 or 0) that the value was found in the list. 我想将其转换为多个列,每个列对应一个不同的值,指示(在1或0中)该值在列表中找到。 Desired result given the above example: 以上示例给出了期望的结果:

df$Ab = 1,0,0,1
df$B = 1,0,0,0
df$C = 1,0,1,0
df$F = 0,0,0,1
df$N = 0,0,0,1

What is the best way to do this? 做这个的最好方式是什么?

Assuming you are starting with: 假设你开始:

df <- data.frame(v1 = c("Ab B C", "", "X C", "N Ab F S"))

You can try cSplit_e from my "splitstackshape" package: 您可以从我的“splitstackshape”包中尝试cSplit_e

library(splitstackshape)
cSplit_e(df, "v1", sep = " ", type = "character", fill = 0)
#         v1 v1_Ab v1_B v1_C v1_F v1_N v1_S v1_X
# 1   Ab B C     1    1    1    0    0    0    0
# 2              0    0    0    0    0    0    0
# 3      X C     0    0    1    0    0    0    1
# 4 N Ab F S     1    0    0    1    1    1    0

You can try 你可以试试

library(qdapTools)
lst <- strsplit(df1$Col1, ' ')
cbind(df1, mtabulate(lst))
#      Col1 Ab B C F N S X
#1   Ab B C  1 1 1 0 0 0 0
#2           0 0 0 0 0 0 0
#3      X C  0 0 1 0 0 0 1
#4 N Ab F S  1 0 0 1 1 1 0

Or using base R 或使用base R

lvls <- sort(unique(unlist(lst)))
cbind(df1, t(vapply(lst, function(x) table(factor(x, levels=lvls)),
                numeric(length(lvls)))))

data 数据

df1 <- structure(list(Col1 = c("Ab B C", "", "X C", "N Ab F S")),
.Names = "Col1", row.names = c(NA, -4L), class = "data.frame")

In base R, another approach: 在基地R,另一种方法:

lst  = strsplit(df$Col1, ' ')
cols = unique(unlist(lst))

m = do.call(rbind, lapply(lst, function(u) cols %in% u +0))
colnames(m) = cols

#> m
#     Ab B C X N F S
#[1,]  1 1 1 0 0 0 0
#[2,]  0 0 0 0 0 0 0
#[3,]  0 0 1 1 0 0 0
#[4,]  1 0 0 0 1 1 1

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM