仅从R中的列中筛选特定字符串

我有一个数据集，其中的列具有逗号分隔的值。我需要解析此列中的每个值，只保留特定值，删除其他值

我拥有的代码和数据如下：

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)

如果列在regionInterest的字符串上有任何内容，我希望保留该字符串，否则我希望删除它

我期望的结果是：

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

---

有人能帮我做错事吗？谢谢

您得到的错误是由于strsplit创建的行数多于您的输入df。同样，在ifelse语句中，您返回的z为FALSE，因此它没有按照您的意图执行

这里有一个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

---

您得到的错误是由于strsplit创建的行比您的输入df多。同样，在ifelse语句中，您返回的z为FALSE，因此它没有按照您的意图执行

这里有一个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

您可能应该首先拆分数据，然后才运行子集。 这将提高效率，因为strsplit是矢量化的，每个分割中的矢量大小无关紧要。而且，在trimws中不需要，它只会使代码效率低下。相反，请在指定fixed=TRUE的同时打开split。这将使strsplit的工作速度提高约X10倍，因为它不会使用正则表达式进行拆分

以下内容仅适用于base 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"

尽管上述解决方案与您自己的解决方案类似，但仍然是一个划行的解决方案。相反，我们可以尝试通过按列操作来实现相同的效果。从列的角度来看，我的意思是，如果我们使用转置拆分，迭代次数将是myDf$GeogPreferences中最长句子的大小。我们拆分的逗号的数量应该大大小于数据中的行数

这是一个使用data.table:：tstrsplit的说明

下面是关于100K行数据集的简单基准测试

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 

因此，其他两个解决方案的运行时间超过了15秒，而我的两个解决方案的运行时间都不到一秒钟。您可能应该首先拆分数据，然后再运行子集。 这将提高效率，因为strsplit是矢量化的，每个分割中的矢量大小无关紧要。而且，在trimws中不需要，它只会使代码效率低下。相反，请在指定fixed=TRUE的同时打开split。这将使strsplit的工作速度提高约X10倍，因为它不会使用正则表达式进行拆分

以下内容仅适用于base 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"

尽管上述解决方案与您自己的解决方案类似，但仍然是一个划行的解决方案。相反，我们可以尝试通过按列操作来实现相同的效果。从列的角度来看，我的意思是，如果我们使用转置拆分，迭代次数将是myDf$GeogPreferences中最长句子的大小。我们拆分的逗号的数量应该大大小于数据中的行数

这是一个使用data.table:：tstrsplit的说明

下面是关于100K行数据集的简单基准测试

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 

---

因此，与我的两个解决方案相比，其他两个解决方案的运行时间都超过了15秒。这两个解决方案的运行时间都不到一秒。如果您喜欢更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 = ", ")

---

如果您喜欢更接近您的方法的解决方案，我希望这会有所帮助，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 = ", ")

---

我希望这有帮助

谢谢杰克！！这很有帮助。谢谢你，杰克！！这很有帮助。谢谢你，马克！！这很有帮助。谢谢你，马克！！这很有帮助。谢谢大卫！！这很有帮助。这是一个很好的比较。谢谢你，大卫。如果需要对现有列进行更改并将其保存回数据集，我需要做什么？即使在使用rbind.data.frame之后，我也会出现此错误。错误：每个变量必须是1d原子向量或列表。问题变量：“地理。首选项”只需重新分配。例如，我在重新分配myDf$GeogPreferences时检查了它是否工作正常。我的代码在另一行有问题。谢谢你的帮助。谢谢大卫！！这很有帮助。这是一个很好的比较。谢谢你，大卫。如果需要对现有列进行更改并将其保存回数据集，我需要做什么？即使在使用rbind.data.frame之后，我也会出现此错误。错误：每个变量必须是1d原子向量或列表。问题变量：“地理。首选项”只需重新分配。例如，我在重新分配myDf$GeogPreferences时检查了它是否工作正常。我的代码在另一行有问题。谢谢你的帮助。