我希望在这个棘手的字符串问题上得到一些帮助。
当前数据框
ID Text
1 This is a very long piece of string. This contains many lines.
我想把它改成：
ID Text1 Text2 Text3 Text4 Text5
1 This is a very long piece of string. This contains many lines.
字符串拆分应该发生在平均拼接数量的单词上。在上面的示例中，我尝试将行平均拆分 5 次，因此每列应包含 20% 的单词。
这背后的目的是将这些词构建成这样一种方式，即它们可以被视为时间序列数据，因为对话刚刚被拆分。
【问题讨论】：
可能有更好的选择，但无需额外的软件包即可：
首先，我们创建一个reproducible example：
df <- data.frame(ID=1:2,
Text=c("Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.",
"Lorem ipsum dolor sit amet, consectetur adipiscing elit"),
stringsAsFactors = FALSE)
然后，chunkize 是 split+cut 的包装，这是棘手的部分。它需要一个character，将其拆分为空格并分成n 块，然后返回一个带有n 多列的data.frame。 （我们删除names，这样rbind 就可以了）。
chunkize <- function(chr, n=5){
x <- strsplit(chr, " ")[[1]]
df <- as.data.frame(
lapply(
split(x,
cut(seq_along(x),
breaks=n)),
paste, collapse=" "),
stringsAsFactors = FALSE, col.names=NULL)
names(df) <- NULL
df
}
然后我们简单地将它应用于每一行。我们还添加了ID 列：
df_chunked <- do.call("rbind",
apply(df, 1,
function(x) cbind(x[1], chunkize(x[-1], 5))))
最后，我们重命名列：
colnames(df_chunked) <- c("ID", paste0("Text", 1:5))
同样的事情变成了一个方便的功能：
chunkize_this <- function(df, n=5){
chunkize <- function(chr, n){
x <- strsplit(chr, " ")[[1]]
df <- as.data.frame(
lapply(
split(x,
cut(seq_along(x),
breaks=n)),
paste, collapse=" "),
stringsAsFactors = FALSE, col.names=NULL)
names(df) <- NULL
df
}
df_chunked <- do.call("rbind",
apply(df, 1, function(x) cbind(x[1], chunkize(x[-1], n))))
colnames(df_chunked) <- c(colnames(df)[1], paste0("Text", 1:n))
rownames(df_chunked) <- NULL
df_chunked
}
你可以试试：
View(chunkize_this(df, 3))
View(chunkize_this(df, 5))
另一个例子：
df <- read.table(h=T, text=
'ID Text
1 "This is a very long piece of string. This contains many lines."
2 "This is a very long piece of string. It contains one or two more word."
3 "Short"'
)
> chunkize_this(df, 5)
ID Text1 Text2 Text3 Text4 Text5
1 1 This is a very long piece of string. This contains many lines.
2 2 This is a very long piece of string. It contains one or two more word.
3 3 Short
【讨论】：
在data.table、base R 和tidyverse 中实现的替代方法。零件数量可以硬编码或预先分配：
# pre-allocating number of parts
np <- 5
不同的选择：
1) 带有“data.table”：
library(data.table)
# method 1
setDT(DF)[, strsplit(Text, "\\s"), by = ID
][, grp := rleid(cut(1:.N, np)), by = ID
][, paste(V1, collapse = " "), by = .(ID, grp)
][, dcast(.SD, ID ~ paste0('Text', grp), value.var = "V1")]
# method 2
setDT(DF)[, strsplit(Text, ' '), by = ID
][, grp := {s <- ceiling(.N/np); rleid(s:(.N+s-1) %/% (.N/np))}, by = ID
][, paste(V1, collapse = ' '), by = .(ID, grp)
][, dcast(.SD, ID ~ paste0('Text', grp), value.var = 'V1')]
两者都给出：
ID Text1 Text2 Text3 Text4 Text5 1: 1 This is a very long piece of string. This contains many lines. 2: 2 This is a very long piece of string. It contains one or two more words. 3: 3 Short text NA NA NA
2) 基础 R：
# method 1
equal_parts <- function(x, np = 5) {
n <- cut(seq_along(x), np)
n <- as.integer(n)
cumsum(c(1, diff(n) > 0))
}
# method 2
equal_parts <- function(x, np = 5) {
n <- length(x)
s <- ceiling(n/np)
rl <- rle(s:(n+s-1) %/% (n/np))$lengths
rep(seq_along(rl), rl)
}
DF.long <- stack(setNames(strsplit(DF$Text, ' '), DF$ID))
DF.long$grp <- with(DF.long, ave(values, ind, FUN = equal_parts))
DF.agg <- aggregate(values ~ ind + grp, DF.long, paste0, collapse = ' ')
reshape(DF.agg, idvar = 'ind', timevar = 'grp', direction = 'wide')
给出：
ind values.1 values.2 values.3 values.4 values.5 1 1 This is a very long piece of string. This contains many lines. 2 2 This is a very long piece of string. It contains one or two more words. 3 3 Short text <NA> <NA> <NA>
3) 'tidyverse'：
library(dplyr)
library(tidyr)
separate_rows(DF, Text) %>%
group_by(ID) %>%
mutate(grp = equal_parts(Text)) %>% # using the 'equal_parts'-function from the base R solution
group_by(grp, add = TRUE) %>%
summarise(Text = paste0(Text, collapse = ' ')) %>%
spread(grp, Text)
给出：
# A tibble: 3 x 6 # Groups: ID [3] ID `1` `2` `3` `4` `5` * <int> <chr> <chr> <chr> <chr> <chr> 1 1 This is a very long piece of string. This contains many lines. 2 2 This is a very long piece of string. It contains one or two more words. 3 3 Short text <NA> <NA> <NA>
使用过的数据：
DF <- structure(list(ID = 1:3, Text = c("This is a very long piece of string. This contains many lines.",
"This is a very long piece of string. It contains one or two more words.",
"Short text")),
.Names = c("ID", "Text"), row.names = c(NA, -3L), class = "data.frame")
【讨论】：
OP 提供了一个只有一行的数据帧。因此，在text 中包含不同数量单词的多行的情况下，尚不清楚预期结果是什么。是否需要
如果要求每列在所有行中都应包含相同数量的单词（如果有足够的单词可用），则单词最多的行将确定分布。单词较少的行的列从左侧开始填充（左对齐）。
library(data.table)
n_brks <- 5L
setDT(DT)[, strsplit(Text, "\\s"), by = ID][
, paste(V1, collapse = " "), by = .(ID, cut(rowid(ID), n_brks))][
, dcast(.SD, ID ~ rowid(ID, prefix = "Text"), fill = "", value.var = "V1")]
ID Text1 Text2 Text3 Text4 Text5 1: 1 This is a very long piece of string. This contains many lines. 2: 2 This is a very long piece of string. It contains one or two more words. 3: 3 Short text 4: 4 Shorter
Text1 到 Text4 列的第 1 行和第 2 行包含相同数量的单词（每个 3 个）。单词数少于列数的行从左侧开始填充。 p>
library(data.table)
DT <- fread( 'ID Text 1 "This is a very long piece of string. This contains many lines." 2 "This is a very long piece of string. It contains one or two more words." 3 "Short text" 4 "Shorter"')
强制转换为 data.table 后，每行中的文本在单词边界处被拆分并以长格式返回（可能被视为等同于时间序列）：
n_brks <- 5L
setDT(DT)[, strsplit(Text, "\\s"), by = ID]
ID V1 1: 1 This 2: 1 is 3: 1 a 4: 1 very 5: 1 long 6: 1 piece 7: 1 of 8: 1 string. 9: 1 This 10: 1 contains 11: 1 many 12: 1 lines. 13: 2 This 14: 2 is 15: 2 a 16: 2 very 17: 2 long 18: 2 piece 19: 2 of 20: 2 string. 21: 2 It 22: 2 contains 23: 2 one 24: 2 or 25: 2 two 26: 2 more 27: 2 words. 28: 3 Short 29: 3 text 30: 4 Shorter ID V1
然后使用计算的分组变量再次连接单词，该变量使用 rowdid() 编号上的 cut() 函数创建 n_brks 块：
setDT(DT)[, strsplit(Text, "\\s"), by = ID][
, paste(V1, collapse = " "), by = .(ID, cut(rowid(ID), n_brks))]
ID cut V1 1: 1 (0.986,3.8] This is a 2: 1 (3.8,6.6] very long piece 3: 1 (6.6,9.4] of string. This 4: 1 (9.4,12.2] contains many lines. 5: 2 (0.986,3.8] This is a 6: 2 (3.8,6.6] very long piece 7: 2 (6.6,9.4] of string. It 8: 2 (9.4,12.2] contains one or 9: 2 (12.2,15] two more words. 10: 3 (0.986,3.8] Short text 11: 4 (0.986,3.8] Shorter
最后，此结果再次从长格式重新调整为宽格式，以创建预期的结果。列标题由rowid() 函数创建，缺失值由"" 替换：
setDT(DT)[, strsplit(Text, "\\s"), by = ID][
, paste(V1, collapse = " "), by = .(ID, cut(rowid(ID), n_brks))][
, dcast(.SD, ID ~ rowid(ID, prefix = "Text"), fill = "", value.var = "V1")]
如果要求将每一行单独拆分并且单词均匀分布，则每列中的单词数将因列而异。单词数少于列数的行每列最多有一个单词。
这种情况的解决方案是修改Jaaps's suggestion：
library(data.table)
n_brks <- 5L
setDT(DT)[, strsplit(Text, "\\s"), by = ID][
, ri := cut(seq_len(.N), n_brks), by = ID][
, paste(V1, collapse = " "), by = .(ID, ri)][
, dcast(.SD, ID ~ rowid(ID, prefix = "Text"), fill = "", value.var = "V1")]
ID Text1 Text2 Text3 Text4 Text5 1: 1 This is a very long piece of string. This contains many lines. 2: 2 This is a very long piece of string. It contains one or two more words. 3: 3 Short text 4: 4 Shorter
现在，每列中的单词数因行而异。例如，Text2 到 Text4 列的第 1 行各有 2 个字，第 2 行各有 3 个字。第 3 行的 2 个字放置在不同的列中。
【讨论】：
setDT(DT)[, strsplit(Text, "\\s"), by = ID][, ri := rowid(ID)][, ri := cut(ri, 5), by = ID][, paste(V1, collapse = " "), by = .(ID, ri)][, dcast(.SD, ID ~ rowid(ID, prefix = "Text"), fill = "", value.var = "V1")]