当R值中的值为NA时，如何跳过一个paste（）参数 [英] How to skip a paste() argument when its value is NA in R

问题描述

我有一个包含城市，州和国家/地区的数据框。我想创建一个连接：城市，州，国家的字符串。但是，我的一个城市没有州（有一个 NA 而已）。我想要这个城市的字符串是城市，国家。以下是创建错误字符串的代码：

I have a data frame with the columns city, state, and country. I want to create a string that concatenates: "City, State, Country". However, one of my cities doesn't have a State (has a NA instead). I want the string for that city to be "City, Country". Here is the code that creates the wrong string:

# define City, State, Country
  city <- c("Austin", "Knoxville", "Salk Lake City", "Prague")
  state <- c("Texas", "Tennessee", "Utah", NA)
  country <- c("United States", "United States", "United States", "Czech Rep")
# create data frame
  dff <- data.frame(city, state, country)
# create full string
  dff["string"] <- paste(city, state, country, sep=", ")

当我显示 dff $ string 时，我得到以下内容。请注意，最后一个字符串有一个 NA，，这是不需要的：

When I display dff$string, I get the following. Note that the last string has a NA,, which is not needed:

> dff["string"]
                               string
1        Austin, Texas, United States
2 Knoxville, Tennessee, United States
3 Salk Lake City, Utah, United States
4               Prague, NA, Czech Rep

我如何跳过 NA ，，包括 sep =，。

推荐答案

替代方法是稍后修复：

The alternative is to just fix it up afterwards:

gsub("NA, ","",dff$string)

#[1] "Austin, Texas, United States"       
#[2] "Knoxville, Tennessee, United States"
#[3] "Salk Lake City, Utah, United States"
#[4] "Prague, Czech Rep"   

另一种方法是使用 data.frame 调用 dff ：

Alternative #2, is to use apply once you have your data.frame called dff:

apply(dff, 1, function(x) paste(na.omit(x),collapse=", ") )

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