提问人：Simon Harmel 提问时间：11/11/2023 最后编辑：Simon Harmel 更新时间：11/17/2023 访问量：81
将文本作为数据帧读入 R
reading text as a dataframe into R
问：
在我的下面，要求列在其元素之间没有任何空格。例如，如果元素是 ，则抛出错误，直到用户删除空格，如 。DATAread.table()study"Hayati & Jalilifar"read.table()"Hayati&Jalilifar"
但是有没有办法阅读以下内容，而无需按原样删除任何数据元素之间的空格？read.table()DATA
DATA = read.table(header=TRUE, text =
"study year g v_g assign_type n_class Nt Nc
Hayati & Jalilifar 2009 0.213 0.101 student NA 20 20
Hayati & Jalilifar 2009 0.785 0.108 student NA 20 20
Hale & Courtney 1994 -0.894 0.0154 class 4 286 286
Hale & Courtney 1994 0.946 0.0156 class 4 286 286
Hale & Courtney 1994 -0.237 0.0146 class 4 277 277
Hale & Courtney 1994 -0.179 0.0146 class 4 277 277")
答：
2赞
stefan
11/11/2023
#1
您的数据似乎是固定宽度的格式。根据您的示例数据，这里是一种几乎完美的方法，除了它将第一列拆分为两列之外，它几乎完美。此外，它需要第二步来获取列名：readr::read_fwf
library(readr)
library(dplyr, warn = FALSE)
tmp <- tempfile()
writeLines(
text = "study year g v_g assign_type n_class Nt Nc
Hayati & Jalilifar 2009 0.213 0.101 student NA 20 20
Hayati & Jalilifar 2009 0.785 0.108 student NA 20 20
Hale & Courtney 1994 -0.894 0.0154 class 4 286 286
Hale & Courtney 1994 0.946 0.0156 class 4 286 286
Hale & Courtney 1994 -0.237 0.0146 class 4 277 277
Hale & Courtney 1994 -0.179 0.0146 class 4 277 277",
tmp
)
dat <- readr::read_fwf(
file = tmp, skip = 1
) |>
mutate(X1 = paste(X1, X2), .keep = "unused")
names(dat) <- readr::read_table(tmp, n_max = 0) |> names()
dat
#> # A tibble: 6 × 8
#> study year g v_g assign_type n_class Nt Nc
#> <chr> <dbl> <dbl> <dbl> <chr> <dbl> <dbl> <dbl>
#> 1 Hayati & Jalilifar 2009 0.213 0.101 student NA 20 20
#> 2 Hayati & Jalilifar 2009 0.785 0.108 student NA 20 20
#> 3 Hale & Courtney 1994 -0.894 0.0154 class 4 286 286
#> 4 Hale & Courtney 1994 0.946 0.0156 class 4 286 286
#> 5 Hale & Courtney 1994 -0.237 0.0146 class 4 277 277
#> 6 Hale & Courtney 1994 -0.179 0.0146 class 4 277 277
1赞
Friede
11/11/2023
#2
我有时使用一种解决方法：base-R
txt <- "study year g v_g assign_type n_class Nt Nc
Hayati & Jalilifar 2009 0.213 0.101 student NA 20 20
Hayati & Jalilifar 2009 0.785 0.108 student NA 20 20
Hale & Courtney 1994 -0.894 0.0154 class 4 286 286
Hale & Courtney 1994 0.946 0.0156 class 4 286 286
Hale & Courtney 1994 -0.237 0.0146 class 4 277 277
Hale & Courtney 1994 -0.179 0.0146 class 4 277 277"
data <- read.table(text = txt, header = FALSE, skip = 1L)
data$V1 <- with(data, paste(V1, V2, V3))
data[, c("V2", "V3")] <- list(NULL)
colnames(data) <- read.table(text = txt, nrows = 1L)
给
> head(data)
study year g v_g assign_type n_class Nt Nc
1 Hayati & Jalilifar 2009 0.213 0.1010 student NA 20 20
2 Hayati & Jalilifar 2009 0.785 0.1080 student NA 20 20
3 Hale & Courtney 1994 -0.894 0.0154 class 4 286 286
4 Hale & Courtney 1994 0.946 0.0156 class 4 286 286
5 Hale & Courtney 1994 -0.237 0.0146 class 4 277 277
6 Hale & Courtney 1994 -0.179 0.0146 class 4 277 277
3赞
Onyambu
11/11/2023
#3
将文本保存在变量中，并使用以下命令：
read.table(text=gsub("(\\S+\\s+[&]\\s+\\S+)", "'\\1'", txt), header = TRUE)
study year g v_g assign_type n_class Nt Nc
1 Hayati & Jalilifar 2009 0.213 0.1010 student NA 20 20
2 Hayati & Jalilifar 2009 0.785 0.1080 student NA 20 20
3 Hale & Courtney 1994 -0.894 0.0154 class 4 286 286
4 Hale & Courtney 1994 0.946 0.0156 class 4 286 286
5 Hale & Courtney 1994 -0.237 0.0146 class 4 277 277
6 Hale & Courtney 1994 -0.179 0.0146 class 4 277 277
txt <- "study year g v_g assign_type n_class Nt Nc
Hayati & Jalilifar 2009 0.213 0.101 student NA 20 20
Hayati & Jalilifar 2009 0.785 0.108 student NA 20 20
Hale & Courtney 1994 -0.894 0.0154 class 4 286 286
Hale & Courtney 1994 0.946 0.0156 class 4 286 286
Hale & Courtney 1994 -0.237 0.0146 class 4 277 277
Hale & Courtney 1994 -0.179 0.0146 class 4 277 277"
评论
1赞
stefan
11/11/2023
真的很好，很优雅。
1赞
SAL
11/17/2023
#4
该函数有两个有用的参数，用于读取此类数据，其中有很多空格和...：它们是和。在本例中，第一个应设置为 。为了防止从基本 R 进行任何自动转换，您还可以添加参数 。这就是您需要做的：read.table()strip.white=sep="\t"TRUEcheck.names=F
DATA = read.table(header=TRUE, check.names = F, strip.white = T, sep = "\t", text =
"study year g v_g assign_type n_class Nt Nc
Hayati & Jalilifar 2009 0.213 0.101 student NA 20 20
Hayati & Jalilifar 2009 0.785 0.108 student NA 20 20
Hale & Courtney 1994 -0.894 0.0154 class 4 286 286
Hale & Courtney 1994 0.946 0.0156 class 4 286 286
Hale & Courtney 1994 -0.237 0.0146 class 4 277 277
Hale & Courtney 1994 -0.179 0.0146 class 4 277 277")
因此，您必须没有错误，这是您可能正在寻找的数据输出：
DATA
study year g v_g assign_type n_class Nt Nc
1 Hayati & Jalilifar 2009 0.213 0.101 student NA 20 20
2 Hayati & Jalilifar 2009 0.785 0.108 student NA 20 20
3 Hale & Courtney 1994 -0.894 0.0154 class 4 286 286
4 Hale & Courtney 1994 0.946 0.0156 class 4 286 286
5 Hale & Courtney 1994 -0.237 0.0146 class 4 277 277
6 Hale & Courtney 1994 -0.179 0.0146 class 4 277 277
评论