Question

structure(list(Other = c(NA_character_, NA_character_, NA_character_,
                         NA_character_, NA_character_),
              Years = c("2005, 2005, 2006, 2006, 2007", "2011, 2014",
                        "2007", "2011, 2011, 2011, 2012, 2012, 2012",
                        "2006, 2006, 2012, 2012, 2015")),
         .Names = c("Other", "Years"), row.names = 1:4, class = "data.frame")

鉴于上述数据框，第二列具有以逗号排列的一堆年份。我想创建一个新列，该列将列中每个元素的年总数相加。因此，最终的数据帧如下所示：

structure(list(Other = c(NA_character_, NA_character_, NA_character_,
                         NA_character_, NA_character_),
               Years = c("2005, 2005, 2006, 2006, 2007","2011, 2014", "2007",
                         "2011, 2011, 2011, 2012, 2012, 2012",
                         "2006, 2006, 2012, 2012, 2015"), 
               yearlength = c(5, 2, 1, 6, 5)),
         .Names = c("Other", "Years", "yearlength"), row.names = 1:4, class = "data.frame")

我尝试使用stack$yearlength <- count.fields(textConnection(stack), sep = ",")之类的解决方案，但无法完全解决问题。

Answer 1

一种方法是计算逗号并添加import { Component, Input, Output, EventEmitter, OnDestroy } from '@angular/core'; @Component({ selector: 'child', template: ` <br><br>I'm a child<br>`, styles: [`h1 { font-family: Lato; }`] }) export class ChildComponent implements OnDestroy { @Output() beingDestroyed = new EventEmitter<boolean>(); ngOnDestroy(): void { this.beingDestroyed.emit(); } }

另一种方法是计算数字的跨度：

df$yearlength <- stringr::str_count(df$Years, ",")+1
df
#output
  Other                              Years yearlength
1  <NA>       2005, 2005, 2006, 2006, 2007          5
2  <NA>                         2011, 2014          2
3  <NA>                               2007          1
4  <NA> 2011, 2011, 2011, 2012, 2012, 2012          6
5  <NA>       2006, 2006, 2012, 2012, 2015          5

第三个选择（要感谢Sotos的评论）是对单词进行计数：

df$yearlength <- stringr::str_count(df$Years, "\\d+")

或

stringi::stri_count_words(df$Years)

第四个选项是计算非空格：

stringr::str_count(df$Years, "\\w+")

编辑：当数据集中不存在NA：

stringr::str_count(df$Years, "\\S+")

all.equal(stringr::str_count(df$Years, ",")+1,
          stringr::str_count(df$Years, "\\d+"),
          stringi::stri_count_words(df$Years),
          stringr::str_count(df$Years, "\\w+"),
          stringr::str_count(df$Years, "\\S+"))

以上所有解决方案产生＃输出 5 2 NA 6 5

将NA更改为0：

df[3,2] <- NA

数据（由于问题中的数据已损坏）：

df$yearlength[is.na(df$yearlength)] <- 0
#output
  Other                              Years yearlength
1  <NA>       2005, 2005, 2006, 2006, 2007          5
2  <NA>                         2011, 2014          2
3  <NA>                               <NA>          0
4  <NA> 2011, 2011, 2011, 2012, 2012, 2012          6
5  <NA>       2006, 2006, 2012, 2012, 2015          5

Answer 2

您可以根据逗号进行分割，然后只需找到向量的长度即可。

> sapply(strsplit(xy$Years, ","), length)
[1] 5 2 1 6 5

已添加到不适用帐户（例如@missuse的帐户）：

xy <- structure(list(Other = c(NA_character_, NA_character_, NA_character_, 
                         NA_character_, NA_character_), Years = c("2005, 2005, 2006, 2006, 2007", 
                                                                  "2011, 2014", "2007", "2011, 2011, 2011, 2012, 2012, 2012", "2006, 2006, 2012, 2012, 2015"
                         )), .Names = c("Other", "Years"), row.names = 1:4, class = "data.frame")

xy[3, 2] <- NA

sapply(strsplit(xy$Years, ","), FUN = function(x) {
  length(na.omit(x))
})

[1] 5 2 0 6 5

对用逗号分隔的字符串总数求和

2 个答案: