合并和替换字符列答案

【问题标题】：Merge and replace character columns合并和替换字符列
【发布时间】：2021-01-16 13:20:45
【问题描述】：

我有一个要合并的包含 2 列的数据框：

     Region             PA
1     Mbeya    Ruaha National Park
2     Mbeya    Ruaha National Park
3     Mbeya    Ruaha National Park
4     Mbeya    Ruaha National Park
5     Mbeya    Ruaha National Park
6     Mbeya    Ruaha National Park
7     Mbeya    NA
8     Mbeya    NA
9     Mbeya    NA
10    Mbeya    NA

这可以通过采用 PA 值并覆盖行中的 Region 值来合并，或者将 PA 中的所有 NA 替换为该行的 Region 中的值。

我试过了：

  Carcass.cleaned$New<-rowSums(Carcass.cleaned[, c("PA", "Region")], na.rm=T)
    Error in base::rowSums(x, na.rm = na.rm, dims = dims, ...) : 
      'x' must be numeric
    
    with(Carcass.cleaned,ifelse(is.na(PA),Region,PA))
    (returns list of numbers)
    
    and coalesce(Carcass.cleaned$PA, Carcass.cleaned$Region) 
    unite(Carcass.cleaned, new, PA:Region, sep='') 
 (both merge the columns names instead of replacing)

【问题讨论】：

标签： r dplyr merge coalesce

【解决方案1】：

您可以使用简单的 if else 语句：

df$Region <- ifelse(is.na(df$PA), df$Region, df$PA)

基本上，只要 PA 为 NA，您就可以保持 Region 不变，而 PA 有值时，您会覆盖 Region 中的值。后记如果你愿意，你可以删除 PA

【讨论】：

【解决方案2】：

尝试使用来自dplyr 的mutate()：

library(tidyr)
library(dplyr)
#Code
df <- df %>% group_by(Region) %>% 
  mutate(PA=ifelse(is.na(PA),Region,PA))

输出：

# A tibble: 10 x 2
# Groups:   Region [1]
   Region PA                 
   <chr>  <chr>              
 1 Mbeya  Ruaha National Park
 2 Mbeya  Ruaha National Park
 3 Mbeya  Ruaha National Park
 4 Mbeya  Ruaha National Park
 5 Mbeya  Ruaha National Park
 6 Mbeya  Ruaha National Park
 7 Mbeya  Mbeya              
 8 Mbeya  Mbeya              
 9 Mbeya  Mbeya              
10 Mbeya  Mbeya

使用的一些数据：

#Data
df <- structure(list(Region = c("Mbeya", "Mbeya", "Mbeya", "Mbeya", 
"Mbeya", "Mbeya", "Mbeya", "Mbeya", "Mbeya", "Mbeya"), PA = c("Ruaha National Park", 
"Ruaha National Park", "Ruaha National Park", "Ruaha National Park", 
"Ruaha National Park", "Ruaha National Park", NA, NA, NA, NA)), row.names = c(NA, 
-10L), class = "data.frame")

【讨论】：

【解决方案3】：

我们可以使用来自dplyr的coalesce

library(dplyr)
df %>%
   mutate(PA = coalesce(PA, Region))
#   Region                  PA
#1   Mbeya Ruaha National Park
#2   Mbeya Ruaha National Park
#3   Mbeya Ruaha National Park
#4   Mbeya Ruaha National Park
#5   Mbeya Ruaha National Park
#6   Mbeya Ruaha National Park
#7   Mbeya               Mbeya
#8   Mbeya               Mbeya
#9   Mbeya               Mbeya
#10  Mbeya               Mbeya

或者在data.table中使用fcoalesce

library(data.table)
setDT(df)[, PA := fcoalesce(PA, Region)]

数据

df <- structure(list(Region = c("Mbeya", "Mbeya", "Mbeya", "Mbeya", 
"Mbeya", "Mbeya", "Mbeya", "Mbeya", "Mbeya", "Mbeya"), PA = c("Ruaha National Park", 
"Ruaha National Park", "Ruaha National Park", "Ruaha National Park", 
"Ruaha National Park", "Ruaha National Park", NA, NA, NA, NA)), row.names = c(NA, 
-10L), class = "data.frame")

【讨论】：

【解决方案4】：

您可以将PA 中的NA 值替换为对应的Region 值。

df$PA[is.na(df$PA)] <- df$Region[is.na(df$PA)]
df
#   Region                  PA
#1   Mbeya Ruaha National Park
#2   Mbeya Ruaha National Park
#3   Mbeya Ruaha National Park
#4   Mbeya Ruaha National Park
#5   Mbeya Ruaha National Park
#6   Mbeya Ruaha National Park
#7   Mbeya               Mbeya
#8   Mbeya               Mbeya
#9   Mbeya               Mbeya
#10  Mbeya               Mbeya

【讨论】：