R：根据 NA 位置合并来自同一 data.frame 的列答案

【问题标题】：R: merge columns from same data.frame based on NA positionsR：根据 NA 位置合并来自同一 data.frame 的列
【发布时间】：2016-11-03 15:12:35
【问题描述】：

我有一个这样的数据框：

df <- data.frame(theme1=c("hello",NA,NA,NA), theme2=c(NA,"world",NA,NA), theme3=c(NA,NA,"good_morning",NA), theme4=c(NA,NA,NA,"good_evening"))

theme1 theme2 theme3 theme4 1 hello NA NA NA 2 NA world NA NA 3 NA NA good_morning NA 4 NA NA NA good_evening

现在我想获得一列并保留行顺序：

**Theme_merged** hello world good_morning good_evening

尝试：

merge_themes <- data.frame(cbind(mycol = na.omit(unlist(data2_tst[18:23]))), stringsAsFactors = F)

上面的代码有效，但不保留行顺序，所以当我想将向量放回原始数据帧时，它不再匹配。

真实数据：

dput(head(data2_tst[18:23], n = 50))
structure(list(Theme1 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, "%Bedrukken%", NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
NA, NA, NA, "%Bedrukken%", NA, NA, NA, NA, NA, NA, NA, NA), Theme2 = c(NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "%Nieuwste|Nieuwe|201[6:7]%", 
"%Nieuwste|Nieuwe|201[6:7]%", "%Nieuwste|Nieuwe|201[6:7]%", NA, 
NA, NA, NA, NA, "%Nieuwste|Nieuwe|201[6:7]%", "%Nieuwste|Nieuwe|201[6:7]%", 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "%Nieuwste|Nieuwe|201[6:7]%", 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, "%Nieuwste|Nieuwe|201[6:7]%", 
"%Nieuwste|Nieuwe|201[6:7]%"), Theme3 = c("%Nodig%", NA, "%Nodig%", 
"%Nodig%", "%Nodig%", NA, NA, "%Nodig%", NA, "%Nodig%", NA, NA, 
NA, NA, "%Nodig%", "%Nodig%", "%Nodig%", NA, NA, NA, NA, NA, 
NA, "%Nodig%", "%Nodig%", NA, NA, "%Nodig%", NA, "%Nodig%", "%Nodig%", 
"%Nodig%", NA, "%Nodig%", "%Nodig%", "%Nodig%", NA, NA, NA, "%Nodig%", 
"%Nodig%", NA, "%Nodig%", NA, "%Nodig%", "%Nodig%", NA, "%Nodig%", 
NA, NA), Theme4 = c(NA, "%Kopen%", NA, NA, NA, "%Kopen%", "%Kopen%", 
NA, "%Kopen%", NA, NA, NA, NA, NA, NA, NA, NA, "%Kopen%", "%Kopen%", 
NA, NA, "%Kopen%", "%Kopen%", NA, NA, "%Kopen%", "%Kopen%", NA, 
"%Kopen%", NA, NA, NA, NA, NA, NA, NA, "%Kopen%", "%Kopen%", 
"%Kopen%", NA, NA, NA, NA, "%Kopen%", NA, NA, "%Kopen%", NA, 
NA, NA), Theme5 = c(NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_), Theme6 = c(NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_, NA_character_, NA_character_, 
NA_character_, NA_character_, NA_character_)), .Names = c("Theme1", 
"Theme2", "Theme3", "Theme4", "Theme5", "Theme6"), row.names = 3:52, class = "data.frame")

【问题讨论】：

@nrussell 你完全正确，对不起，这是一个错字！
(1) 请去掉cbind。 (2) 在哪些情况下na.omit(unlist(df, use.names = FALSE)) 不保留订单？我无法重现该问题。总是按预期为我工作。
如果我使用真实数据，也许它会起作用。我会加广告的。所有这些都改变了我机器中按字母顺序排列的行顺序。

标签： r merge cbind

【解决方案1】：

dplyr 0.5.0 版本引入了合并功能：

这个版本的 dplyr 获得了许多受 SQL 启发的向量函数。有两个函数可以更轻松地消除或生成缺失值：

给定一组向量，coalesce() 在每个位置找到第一个非缺失值。

要将其应用于您可以使用的示例数据框：

df <- mutate_all(df, .funs = as.character)
df$merged <- with(df, coalesce(theme1, theme2, theme3, theme4))

我发现有必要从因子转换为字符以避免“无效因子水平”错误。

您的真实数据无需转换：

df$merged <- with(df, coalesce(Theme1, Theme2, Theme3, Theme4, Theme5, Theme6)

【讨论】：

【解决方案2】：

在 SQL 中，这将是 COALESCE 函数：

apply(df, 1, function(r) c(na.omit(r), NA)[1])
# [1] "hello"        "world"        "good_morning" "good_evening"

df <- data.frame(
    theme1=c("hello",NA,NA,NA), 
    theme2=c(NA,"world",NA,NA), 
    theme3=c(NA,NA,"good_morning",NA), 
    theme4=c(NA,NA,NA,"good_evening"),
    stringsAsFactors = FALSE
)

在您的示例数据上na.omit(unlist(df2, use.names = FALSE)) 可以正常工作，但如果有一行 only NA 值，它将失败：

df2 <- data.frame(
    theme1=c("hello",NA,NA,NA,NA), 
    theme2=c(NA,"world",NA,NA,NA), 
    theme3=c(NA,NA,"good_morning",NA,NA), 
    theme4=c(NA,NA,NA,"good_evening",NA),
    theme5=c(NA_character_,NA_character_,NA_character_,
             NA_character_,NA_character_),
    stringsAsFactors = FALSE
)

df2$X <- na.omit(unlist(df2, use.names = FALSE))
# Error in `$<-.data.frame`(`*tmp*`, "X", value = c("hello", "world", "good_morning",  : 
#   replacement has 4 rows, data has 5

df2$X <- apply(df2, 1, function(r) c(na.omit(r), NA)[1])
#   theme1 theme2       theme3       theme4 theme5            X
# 1  hello   <NA>         <NA>         <NA>   <NA>        hello
# 2   <NA>  world         <NA>         <NA>   <NA>        world
# 3   <NA>   <NA> good_morning         <NA>   <NA> good_morning
# 4   <NA>   <NA>         <NA> good_evening   <NA> good_evening
# 5   <NA>   <NA>         <NA>         <NA>   <NA>         <NA>

另一个选项可能是df2$X <- df2[cbind(1:nrow(df2), max.col(!is.na(df2)))]

【讨论】：

@DavidArenburg 我想摆脱主题列，只创建一个合并列。
@DavidArenburg 当我想将向量放回它不再匹配的原始数据帧时，我使用了 "[...]。" 表示结果应该长度与输入数据中的行数相同。
您的回答很好，非常感谢！ @DavidArenburg

【解决方案3】：

这是一个 tidyverse 解决方案（使用 dplyr 和 tidyr 或仅使用 tidyverse）

library(tidyverse)

> df <- df %>% 
    gather("theme", "theme_merged", 1:4) %>%
    filter(!is.na(theme_merged)) %>% 
    select(theme_merged)

> df
  theme_merged
1        hello
2        world
3 good_morning
4 good_evening

【讨论】：

感谢您的回答，但在我的真实数据中，订单并未保留。
它是根据其他变量排序的吗？还是按字母顺序？还是它是有序级别的一个因素？
应该保留订单，因为这只是我的大 DF 的一小部分。我需要将合并的列添加回 DF

【解决方案4】：

这应该适用于您的数据：

new_df = c(as.matrix(df))

此行首先将df 转换为矩阵，并将一个向量中的所有列与c() 绑定。

new_df <- new_df[!is.na(new_df)]

现在我们只保留非NA 条目。如果您愿意，可以将其转换回数据框：

new_df <- data.frame(new_df);names(new_df) <- "Themes"

【讨论】：