我正在尝试用新行“\ n”拆分列中的字符串。 这是一个dataframe sample_data：
test_data <- data.frame(ID=c('john@xxx.com', 'sally@xxx.com'),
Changes=c('3 max cost changes
productxyz > pb100 > a : Max cost decreased from $0.98 to $0.83
productxyz > pb2 > a : Max cost decreased from $1.07 to $0.91
productxyz > pb2 > b : Max cost decreased from $0.65 to $0.55',
'2 max cost changes
productabc > pb1000 > d : Max cost decreased from $1.07 to $0.91
productabc > pb1000 > x : Max cost decreased from $1.44 to $1.22'), stringsAsFactors=FALSE)
我的目标是将价格提取到列中并获得如下结果集：
ID Prev_Price New_Price
john@xxx.com $0.98 $0.83
john@xxx.com $1.07 $0.91
john@xxx.com $0.65 $0.55
sally@xxx.com $1.07 $0.91
sally@xxx.com $1.44 $1.22
我尝试过使用tidyr包，但结果却充满了N / A.
vars <- c("Prev_Price","New_Price")
seperate(sample_data, Changes, into = vars, sep = "[A-Za-z]+from", extra= "drop")
非常感谢任何帮助。
谢谢！
答案 0 :(得分：3)
尝试
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或者
df1$ID <- df1$ID[df1$ID!=''][cumsum(df1$ID!='')]
library(stringi)
setNames(data.frame(df1$ID, do.call(rbind,stri_extract_all(df1$Changes,
regex='\\$\\d*'))), c('ID', 'Prev_Price', 'New_Price'))
# ID Prev_Price New_Price
#1 A $20 $10
#2 A $11 $10
#3 B $13 $12
#4 B $15 $12
或者
library(tidyr)
extract(df1, Changes, into=c('Prev_Price', 'New_Price'),
'[^$]*(\\$\\d*)[^$]*(\\$\\d*)')
# ID Prev_Price New_Price
#1 A $20 $10
#2 A $11 $10
#3 B $13 $12
#4 B $15 $12
注意：＆＃34;更改＆＃34;可以删除
或仅使用library(data.table)#v1.9.5+
setDT(df1)[, c('Prev_Price', 'New_Price') := tstrsplit(Changes,
'[A-Za-z ]+')[-1]][]
# ID Changes Prev_Price New_Price
#1: A down from $20 to $10 $20 $10
#2: A down from $11 to $10 $11 $10
#3: B down from $13 to $12 $13 $12
#4: B down from $15 to $12 $15 $12
方法
base R如果元素位于同一个单元格中，则一个选项将使用devel版本data.frame(ID=df1$ID, read.table(text=gsub('[^$]*(\\$\\d+)', ' \\1 ',
df1$Changes),col.names=c('Prev_Price', 'New_Price'),
stringsAsFactors=FALSE))
# ID Prev_Price New_Price
#1 A $20 $10
#2 A $11 $10
#3 B $13 $12
#4 B $15 $12
即。 v1.9.5 +。它可以从here
在这里，我们使用相同的代码来拆分＆＃39;更改＆＃39; （data.table），然后tstrsplit(Changes,..)将输出设置为长格式，将melt指定为measure.vars，如果需要，list按ID＆＃ 39;并删除不需要的列（＆＃39;变量＆＃39;）。
order或者我们可以像以前一样使用 melt(
setDT(df2)[, paste0('V',1:4) := tstrsplit(Changes,
'[A-Za-z ]+')[-1]][,-2, with=FALSE],
id.var='ID', measure=list(c('V1', 'V3'), c('V2', 'V4')),
value.name=c('Prev_Price', 'New_Price'))[order(ID)][, variable:=NULL]
# ID Prev_Price New_Price
#1: A $20 $10
#2: A $11 $10
#3: B $13 $12
#4: B $15 $12
，然后使用gsub中的long转换为reshape格式
base R对于新数据集（＆＃34; df3＆＃34;），我们可以使用 d1 <- data.frame(ID=df2$ID,read.table(text=gsub('[^$]*(\\$\\d+)',
' \\1 ', df2$Changes)))
colnames(d1)[-1] <- paste0(c('Prev_Price.', 'New_Price.'),
rep(1:2,each=2))
reshape(d1, idvar='ID', varying=2:ncol(d1), sep=".", direction='long')
# ID time Prev_Price New_Price
#A.1 A 1 $20 $10
#B.1 B 1 $13 $12
#A.2 A 2 $11 $10
#B.2 B 2 $15 $12
提取stri_extract_all_regex后跟数字，包括＆的小数（$） ＃34;变更＆＃34;列，使用'\\$[0-9.]+'将第一列与我们在将输出更改为Map后从list获得的stri_extract_all_regex输出结合起来（因为我们需要交替元素为在不同的列中），然后matrix（rbind）。
do.call(rbind,library(stringi)
res <- do.call(rbind,
Map(function(x,y) data.frame(x,matrix(y, ncol=2, byrow=TRUE,
dimnames=list(NULL, c("Prev_Price", "New_Price")))),
df3$ID, stri_extract_all_regex(df3$Changes, '\\$[0-9.]+')))
row.names(res) <- NULL
res
# x Prev_Price New_Price
#1 john@xxx.com $0.98 $0.83
#2 john@xxx.com $1.07 $0.91
#3 john@xxx.com $0.65 $0.55
#4 sally@xxx.com $1.07 $0.91
#5 sally@xxx.com $1.44 $1.22
答案 1 :(得分：1)
df <- data.frame(ID=c('A','','B',''), Changes=c('down from $20 to $10','down from $11 to $10','down from $13 to $12','down from $15 to $12'), stringsAsFactors=F );
with(list(ss=strsplit(df$Changes,'\\s+')),transform(df,ID=ID[ID!=''][cumsum(ID!='')],Prev_Price=sapply(ss,function(v)v[3]),New_Price=sapply(ss,function(v)v[5]),Changes=NULL));
## ID Prev_Price New_Price
## 1 A $20 $10
## 2 A $11 $10
## 3 B $13 $12
## 4 B $15 $12
另一种方法：
with(df,cbind(ID=ID[ID!=''][cumsum(ID!='')],setNames(as.data.frame(do.call(rbind,strsplit(Changes,'\\s+'))[,c(3,5)]),c('Prev_Price','New_Price'))));
## same result