Question

我有一个包含逗号分隔值的列的数据集。我需要解析此列中的每个值，并仅保留特定值并删除其他值。

我的代码和数据是：

myDf <- structure(list(GeogPreferences = structure(1:4, .Label = c("Central and East Europe, Europe, North America, West Europe, US", 
"Europe, North America, West Europe, US", "Global, North America", 
"Northeast, Southeast, West, US"), class = "factor")), .Names = "GeogPreferences", class = "data.frame", row.names = c(NA, 
-4L))

 regionInterest <- c("Americas", "North America", "US", "Northeast","Southeast","West","Midwest","Southwest")
 k<-lapply(as.character(myDf$GeogPreferences),function(x) {
   z<-trimws(unlist(strsplit(x,split = ",")))
   z <- ifelse((z %in% regionInterest), z[z %in% regionInterest], z)
 })
 myDf$GeogPreferences<-unlist(k)

这是我得到的错误：

Error in `$<-.data.frame`(`*tmp*`, "GeogPreferences", value = c("Central and East Europe",
: replacement has 15 rows, data has 4

我的数据集如下所示：

    GeogPreferences
1   Central and East Europe, Europe, North America, West Europe, US
2   Europe, North America, West Europe, US
3   Global, North America
4   Northeast, Southeast, West, US

如果列中有regionInterest的字符串，我想保留该字符串，否则我想删除它。

我期待的输出是：

    GeogPreferences
1   North America, US
2   North America, US
3   North America
4   Northeast, Southeast, West, US

有人可以帮我解决我做错的事吗？谢谢！

Answer 1

您获得的错误是由strsplit创建的行数多于输入df。同样在ifelse声明中，您在FALSE上返回z，因此它没有按照您的意图行事。

以下是您问题的tidyr / dplyr解决方案。

myDf %>% 
  mutate(id = row_number()) %>% 
  separate_rows(GeogPreferences, sep = ",") %>% 
  mutate(GeogPreferences = trimws(GeogPreferences)) %>% 
  filter(GeogPreferences %in% c("Americas", "North America", "US", "Northeast","Southeast","West","Midwest","Southwest")) %>% 
  group_by(id) %>% 
  summarize(GeogPreferences = toString(trimws(GeogPreferences))) %>% 
  select(-id) 


# A tibble: 4 × 1
                 GeogPreferences
                           <chr>
1              North America, US
2              North America, US
3                  North America
4 Northeast, Southeast, West, US

Answer 2

您可能首先拆分数据，然后运行子集。这将提高效率（因为strsplit它是矢量化的）并且每个分裂中的矢量大小都不重要。此外，在trimws中不需要它只会使您的代码效率低下。相反，在指定", "时拆分fixed = TRUE。这将使strsplit的工作速度提高X10倍，因为它不会使用正则表达式进行拆分。

以下作品仅使用基础R

do.call(rbind, # you can use `rbind.data.frame` instead if you don't want a matrix
        lapply(strsplit(as.character(myDf$GeogPreferences), ", ", fixed = TRUE), 
               function(x) toString(x[x %in% regionInterest])))
#     [,1]                            
# [1,] "North America, US"             
# [2,] "North America, US"             
# [3,] "North America"                 
# [4,] "Northeast, Southeast, West, US"

虽然上述解决方案（与您自己类似）仍然是行式解决方案。相反，我们可以尝试通过逐列操作来实现相同的效果。并且通过＆＃34; columnwise＆＃34;我的意思是，如果我们将对转置分割进行操作，迭代次数将是myDf$GeogPreferences中最长句子的大小（我们分割的逗号的数量），它应该明显小于数据中的行。

她是使用data.table::tstrsplit

的插图

tmp <- data.table::tstrsplit(myDf$GeogPreferences, ", ", fixed = TRUE)
res <- do.call(paste, 
               c(sep = ", ", 
                 lapply(tmp, function(x) replace(x, !x %in% regionInterest, NA_character_))))
gsub("NA, |, NA", "", res)
# [1] "North America, US" "North America, US" "North America" "Northeast, Southeast, West, US"

以下是100K行数据集的简单基准

bigDF <- myDf[sample(nrow(myDf), 1e5, replace = TRUE),, drop = FALSE]

library(dplyr)
library(tidyr)
library(data.table)

tidyverse <- function(x) {
  x %>% 
    mutate(id = row_number()) %>% 
    separate_rows(GeogPreferences, sep = ",") %>% 
    mutate(GeogPreferences = trimws(GeogPreferences)) %>% 
    filter(GeogPreferences %in% c("Americas", "North America", "US", "Northeast","Southeast","West","Midwest","Southwest")) %>% 
    group_by(id) %>% 
    summarize(GeogPreferences = toString(trimws(GeogPreferences))) %>% 
    select(-id) 

}

MF <- function(x) {
    k <- lapply(as.character(x$GeogPreferences), function(x) {
      z <- trimws(unlist(strsplit(x, split = ",")))
      z <- z[z %in% regionInterest] 
    })
    sapply(k, paste, collapse = ", ")
}


DA1 <- function(x) {
  do.call(rbind,
          lapply(strsplit(as.character(x$GeogPreferences), ", ", fixed = TRUE), 
                 function(x) toString(x[x %in% regionInterest])))
}

DA2 <- function(x) {
  tmp <- data.table::tstrsplit(x$GeogPreferences, ", ", fixed = TRUE)
  res <- do.call(paste, 
                 c(sep = ", ", 
                   lapply(tmp, function(x) replace(x, !x %in% regionInterest, NA_character_))))
  gsub("NA, |, NA", "", res)
}

system.time(tidyverse(bigDF))
# user  system elapsed 
# 17.67    0.01   17.91 
system.time(MF(bigDF))
# user  system elapsed 
# 15.52    0.00   15.70 
system.time(DA1(bigDF))
# user  system elapsed 
# 0.97    0.00    1.00 
system.time(DA2(bigDF))
# user  system elapsed 
# 0.25    0.00    0.25

所以其他两个解决方案运行时间超过15秒，而我的两个解决方案运行时间不到一秒

Answer 3

如果您更喜欢接近您的方法的解决方案，请将其更改为

regionInterest <- c("Americas", "North America", "US",
  "Northeast","Southeast","West","Midwest","Southwest")
k<-lapply(as.character(myDf$GeogPreferences),function(x) {
  z<-trimws(unlist(strsplit(x,split = ",")))
# this makes sure you only use z which are in regionInterest
  z <- z[z %in% regionInterest] 
})
# paste with collapse creates one value out of a vector of string seperated by the collapse argument
myDf$GeogPreferences<-sapply(k, paste, collapse = ", ")

我希望这会有所帮助

仅过滤R

3 个答案: