Question

我想遍历数据框的列名，然后使用 dplyr ，使用在行字段之间找到的定界符（->）分隔字段。这就是数据集的样子：

dput(df)
structure(list(v1 = c("Silva->Mark", "Brandon->Livo", "Mango->Apple"),
               v2 = c("Austin",  "NA ", "Orange"),
               v3 = c("James -> Jacy","NA->Jane", "apple -> Orange")),
          class = "data.frame", row.names = c(NA,  -3L))

现在，我编写了一个代码，用列v1 和列v3 的行用定界符（->）过滤了列名。这是代码：

rows_true <- apply(df,2,function(x) any(sapply(x,function(y)grepl("->",y))))
ss<-df[,rows_true]

然后我尝试遍历这些列名，以便可以使用分隔符使用此代码进行分隔，但它不起作用

cols<- names(df)

if (names %in% df){
splitcols <- ss %>%
  tidyr::separate(cols, into = c(paste0(names,+ "old"), "paste0(names,+ "New")"), sep = "->")
}

我使用 paste0 的原因是，我确实希望使用定界符将列分成两部分，然后应使用原始名称加后缀来命名新形成的列旧（第一个拆分列），新（第二个拆分列）

遍历列名并递归分离它们后的最终结果应如下图所示

dput(df)
structure(list(v1_Old = c("Silva", "Brandon", "Mango"),
               v1_New = c("Mark", "Livo", "Apple"),
               v3_Old = c("James","NA", "apple"),
               v3_New = c("Jacy","Jane", "Orange")),
          class = "data.frame", row.names = c(NA,  -3L))

Answer 1

涉及dplyr和tidyr的一种可能性是：

df %>%
 select(v1, v3) %>%
 rowid_to_column() %>%
 gather(var, val, -rowid) %>%
 separate_rows(val, sep = "->", convert = TRUE) %>%
 group_by(rowid) %>%
 mutate(val = trimws(val),
        var = make.unique(var)) %>%
 ungroup() %>%
 spread(var, val) %>%
 select(-rowid)

  v1      v1.1  v3    v3.1  
  <chr>   <chr> <chr> <chr> 
1 Silva   Mark  James Jacy  
2 Brandon Livo  <NA>  Jane  
3 Mango   Apple apple Orange

或者进一步匹配预期输出：

df %>%
 select(v1, v3) %>%
 rowid_to_column() %>%
 gather(var, val, -rowid) %>%
 separate_rows(val, sep = "->", convert = TRUE) %>%
 group_by(rowid, var) %>%
 mutate(val = trimws(val),
        var2 = if_else(row_number() == 2, paste0(var, "_old"), paste0(var, "_new"))) %>%
 ungroup() %>%
 select(-var) %>%
 spread(var2, val) %>%
 select(-rowid)

  v1_new  v1_old v3_new v3_old
  <chr>   <chr>  <chr>  <chr> 
1 Silva   Mark   James  Jacy  
2 Brandon Livo   <NA>   Jane  
3 Mango   Apple  apple  Orange

Answer 2

以下是使用dplyr，purr和stringr的另一种方法。

library(dplyr)
library(purrr)
library(stringr)

# Detect the columns with at least on "->"
my_df_cols <- map_lgl(my_df, ~ any(str_detect(., "->")))

my_df %>% 
  # Select only the columns with at least "->"
  select(which(my_df_cols)) %>% 
  # Mutate these columns and only keep the mutated columns with new names
  transmute_all(list(old = ~ str_split(., "->", simplify = TRUE)[, 1], 
                     new = ~ str_split(., "->", simplify = TRUE)[, 2]))

#    v1_old v3_old v1_new  v3_new
# 1   Silva James    Mark    Jacy
# 2 Brandon     NA   Livo    Jane
# 3   Mango apple   Apple  Orange

Answer 3

我们还可以使用cSplit中的splitstackshape

#Detect columns with "->"
cols <- names(df)[colSums(sapply(df, grepl,  pattern = "->")) > 1]
#Remove unwanted whitespaces before and after "->"
df[cols] <- lapply(df[cols],  function(x) gsub("\\s+", "", x))

#Split into new columns specifying sep as "->"
splitstackshape::cSplit(df[cols], cols, sep = "->")

#      v1_1  v1_2  v3_1   v3_2
#1:   Silva  Mark James   Jacy
#2: Brandon  Livo  <NA>   Jane
#3:   Mango Apple apple Orange

Answer 4

为了完整起见，这也是使用data.table()的解决方案。

与到目前为止发布的其他答案有些不同：

不需要预先标识要拆分的列。取而代之的是，将没有"->"的列从结果即时删除。
用于拆分的正则表达式包括周围的空白（如果有）
" *-> *"。这样可以避免以后在生成的图块上调用trimws()或事先删除空白。

。

library(data.table)
library(magrittr)   # piping used to improve readability
setDT(df)
lapply(names(df), function(x) {
  mDT <- df[, tstrsplit(get(x), " *-> *")]
  if (ncol(mDT) == 2L) setnames(mDT, paste0(x, c("_Old", "_New")))
}) %>% as.data.table()

    v1_Old v1_New v3_Old v3_New
1:   Silva   Mark  James   Jacy
2: Brandon   Livo     NA   Jane
3:   Mango  Apple  apple Orange

使用dplyr

4 个答案: