如何整齐地组合稀疏列答案

【问题标题】：how to combine sparse columns tidily如何整齐地组合稀疏列
【发布时间】：2017-04-13 20:45:15
【问题描述】：

一位同事有一些由许多稀疏列组成的数据，应该折叠成几个填充的列。例如：

d1 <- data.frame(X1 = c(rep("Northampton", times=3), rep(NA, times=7)), 
                 X2 = c(rep(NA, times=3), rep("Amherst", times=5), rep(NA, times=2)), 
                 X3 = c(rep(NA, times=8), rep("Hadley", times=2)), 
                 X4 = c(rep("Stop and Shop", times=2), rep(NA, times=6), rep("Stop and Shop", times=2)), 
                 X5 = c(rep(NA, times=2), rep("Whole Foods", times=6), rep(NA, times=2)))

d1
            X1      X2     X3            X4          X5
1  Northampton    <NA>   <NA> Stop and Shop        <NA>
2  Northampton    <NA>   <NA> Stop and Shop        <NA>
3  Northampton    <NA>   <NA>          <NA> Whole Foods
4         <NA> Amherst   <NA>          <NA> Whole Foods
5         <NA> Amherst   <NA>          <NA> Whole Foods
6         <NA> Amherst   <NA>          <NA> Whole Foods
7         <NA> Amherst   <NA>          <NA> Whole Foods
8         <NA> Amherst   <NA>          <NA> Whole Foods
9         <NA>    <NA> Hadley Stop and Shop        <NA>
10        <NA>    <NA> Hadley Stop and Shop        <NA>

X1:X3 应折叠到名为 Town 的一列中，X4:X5 应折叠到名为 Store 的一列中。这里必须有一个 tidyverse 解决方案。我试过gather() 和unite()，但没有发现任何优雅的东西。

【问题讨论】：

标签： r dplyr tidyr tidyverse

【解决方案1】：

你可以使用coalesce:

d1 %>% mutate_if(is.factor, as.character) %>%    # coerce explicitly
    transmute(town = coalesce(X1, X2, X3), 
              store = coalesce(X4, X5))

##           town         store
## 1  Northampton Stop and Shop
## 2  Northampton Stop and Shop
## 3  Northampton   Whole Foods
## 4      Amherst   Whole Foods
## 5      Amherst   Whole Foods
## 6      Amherst   Whole Foods
## 7      Amherst   Whole Foods
## 8      Amherst   Whole Foods
## 9       Hadley Stop and Shop
## 10      Hadley Stop and Shop

【讨论】：

【解决方案2】：

我认为一系列gather() 调用和一些修剪会让你得到你想要的。一个问题是使用na.rm = TRUE 到gather() 的参数来剔除不需要的行。

d1 %>% 
  gather(key = "town", value = "town_name", X1:X3, na.rm = TRUE) %>% 
  gather(key = "store", value = "store_name", X4:X5, na.rm = TRUE) %>%
  select(-town, -store)

这样可以吗？

【讨论】：

【解决方案3】：

您也可以在 base R 中使用apply run rowwise 执行此操作：

d2 <- data.frame(X1 = apply(d1[,c("X1", "X2", "X3")], 1, function(x) x[!is.na(x)]),
                 X2 = apply(d1[,c("X4", "X5")], 1, function(x) x[!is.na(x)]),
                 stringsAsFactors = FALSE)

结果：

> d2
            X1            X2
1  Northampton Stop and Shop
2  Northampton Stop and Shop
3  Northampton   Whole Foods
4      Amherst   Whole Foods
5      Amherst   Whole Foods
6      Amherst   Whole Foods
7      Amherst   Whole Foods
8      Amherst   Whole Foods
9       Hadley Stop and Shop
10      Hadley Stop and Shop

【讨论】：

【解决方案4】：

这是base R 使用pmax/pmin 的另一种方式

data.frame(lapply(list(Town = d1[1:3], Store = d1[4:5]), function(x) 
           do.call(pmax, c(x, na.rm = TRUE))), stringsAsFactors=FALSE)
#          Town         Store
#1  Northampton Stop and Shop
#2  Northampton Stop and Shop
#3  Northampton   Whole Foods
#4      Amherst   Whole Foods
#5      Amherst   Whole Foods
#6      Amherst   Whole Foods
#7      Amherst   Whole Foods
#8      Amherst   Whole Foods
#9       Hadley Stop and Shop
#10      Hadley Stop and Shop

数据

d1 <- data.frame(X1 = c(rep("Northampton", times=3),rep(NA, times=7)),
   X2 = c(rep(NA, times=3), rep("Amherst", times=5), rep(NA, times=2)),
  X3 = c(rep(NA, times=8), rep("Hadley", times=2)), 
  X4 = c(rep("Stop and Shop", times=2), rep(NA, times=6), rep("Stop and Shop", times=2)), 
  X5 = c(rep(NA, times=2), rep("Whole Foods", times=6), 
        rep(NA, times=2)), stringsAsFactors=FALSE)

【讨论】：