拆分具有不同行数的列，包括标题

Question

我有如下数据框

ID  age type_location          value_1  value_2    
1   83  country:province:city  X:A:J    X:A:I
2   15  country:city           X:K      X:J
3   2   country:province:city  Y:B:I    Y:B:I
4   12  country:city           X:L      Z:K
5   2   country:city           Y:J      X:J
6   2   country:province:city  Y:A:M    Y:B:I
7   18  country:province:city  X:B:J    X:A:L
8   85  country:province:city  X:A:I    Y:B:M

描述一下：第三列（type_location），其中记录包含以“：”分隔的字符串，长度不同，这是第四列和第五列中的值的名称。

因此，我需要这样获取data.frame：

ID  age value_1_country value_1_province value_1_city value_2_country value_2_province value_2_city
1   83  X               A                J            X               A                I
2   15  X               'NA'             K            X               'NA'             J
3   2   Y               B                I            Y               B                I
4   12  X               'NA'             L            Z               'NA'             K
5   2   Y               'NA'             J            X               'NA'             J
6   2   Y               A                M            Y               B                I
7   18  X               B                J            X               A                L
8   85  X               A                I            Y               B                M

描述 - 我需要将第 4 列和第 5 列中的值分开，放入新列并根据第 3 列中的记录给它们命名。在第三列中有不同数量的值用“：”分隔。

我应该怎么做才能传播值并保留第四列和第五列的名称？

Answer 1

如果您使用tidyverse（也可以导入tidyr），您可以使用separate：

library(tidyverse)

new_df = df %>%
  separate('value_1', c('value_1_country', 'value_1_province', 'value_1_city'), sep = ":") %>%
  separate('value_2', c('value_2_country', 'value_2_province', 'value_2_city'), sep = ":")

new_df

我从那一步得到的结果：

> new_df
  ID age         type_location value_1_country value_1_province value_1_city value_2_country value_2_province value_2_city
1  1  83 country:province:city               X                A            J               X                A            I
2  2  15          country:city               X                K         <NA>               X                J         <NA>
3  3   2 country:province:city               Y                B            I               Y                B            I
4  4  12          country:city               X                L         <NA>               Z                K         <NA>
5  5   2          country:city               Y                J         <NA>               X                J         <NA>
6  6   2 country:province:city               Y                A            M               Y                B            I
7  7  18 country:province:city               X                B            J               X                A            L
8  8  85 country:province:city               X                A            I               Y                B            M

对于分隔，第一个参数是您的数据框，第二个是要分隔的列，第三个是列名的向量，sep='' 是要分隔的字符。

您还可以指定 NAN 处理和其他内容，请参见此处：

https://tidyr.tidyverse.org/reference/separate.html

另一个人更快地得到了完整的答案，所以我不会完成输入完整的解决方案。

Answer 2

我敢肯定，有比我更优雅的方法，但使用 dplyr/tidyverse 你可以做到以下几点。

将您的数据作为dput(df) 的输出提供总是被认为是礼貌的，因为这可以更容易地读入R。

df=structure(list(ID = 1:8, 
                  age = c(83L, 15L, 2L, 12L, 2L, 2L, 18L, 85L), 
                  type_location = c("country:province:city", "country:city", 
                                    "country:province:city", "country:city",
                                    "country:city", "country:province:city", 
                                    "country:province:city", "country:province:city"),
                  value_1 = c("X:A:J","X:K", "Y:B:I", "X:L", "Y:J", "Y:A:M", "X:B:J", "X:A:I"), 
                  value_2 = c("X:A:I", "X:J", "Y:B:I", "Z:K", "X:J", "Y:B:I", "X:A:L", "Y:B:M")),
             class = "data.frame", row.names = c(NA, -8L))

library(tidyverse)

df_sep=df %>% 
  separate(col=value_1,into=c(paste0("value_1_",c("country","province","city"))),sep=":",remove = F) %>% 
  mutate(value_1_city=ifelse(nchar(value_1)<4,value_1_province,value_1_city),
         value_1_province=ifelse(nchar(value_1)<4,NA,value_1_province)) %>% 
  separate(col=value_2,into=c(paste0("value_2_",c("country","province","city"))),sep=":",remove = F) %>% 
  mutate(value_2_city=ifelse(nchar(value_2)<4,value_2_province,value_2_city),
         value_2_province=ifelse(nchar(value_2)<4,NA,value_2_province))

我设置了remove=FALSE，以便您可以更轻松地检查输出。如果要删除原始列，只需设置remove=T（默认值）。

Answer 3

这是一种 data.table 方法（可能可以缩短）...

library( data.table )

dt <- fread("ID  age type_location          value_1  value_2    
1   83  country:province:city  X:A:J    X:A:I
2   15  country:city           X:K      X:J
3   2   country:province:city  Y:B:I    Y:B:I
4   12  country:city           X:L      Z:K
5   2   country:city           Y:J      X:J
6   2   country:province:city  Y:A:M    Y:B:I
7   18  country:province:city  X:B:J    X:A:L
8   85  country:province:city  X:A:I    Y:B:M")

#get #types of location
z <- length(unique(unlist(strsplit(dt$type_location, ":"))))
#create new columns
dt[, paste0( "location_",1:z) := tstrsplit( type_location, ":" )]
dt[, paste0( "value_1_",1:z) := tstrsplit( value_1, ":" )]
dt[, paste0( "value_2_",1:z) := tstrsplit( value_2, ":" )]
#melt
ans <- melt( dt[, -(3:5)], 
             id.vars = c("ID", "age"), 
             measure.vars = patterns(location = "^location", 
                                     value_1 = "^value_1", 
                                     value_2 = "^value_2") )
#remove rows with NA
ans <- ans[ complete.cases(ans), ]
#cast to desired format
dcast(ans, ID + age ~ location, value.var = c("value_1", "value_2"))

输出

#    ID age value_1_city value_1_country value_1_province value_2_city value_2_country value_2_province
# 1:  1  83            J               X                A            I               X                A
# 2:  2  15            K               X             <NA>            J               X             <NA>
# 3:  3   2            I               Y                B            I               Y                B
# 4:  4  12            L               X             <NA>            K               Z             <NA>
# 5:  5   2            J               Y             <NA>            J               X             <NA>
# 6:  6   2            M               Y                A            I               Y                B
# 7:  7  18            J               X                B            L               X                A
# 8:  8  85            I               X                A            M               Y                B

Answer 4

当然我会选择Ryan's answer，但如果有人想避免硬编码名称，这行得通。

library(tidyr)
library(dplyr)
library(purrr)

df1 %>% 
  group_split(ID) %>% 
  map(., ~ separate_rows(.x, -c(ID, age))) %>% 
  map(., ~ mutate(.x, type_location2 = type_location, 
                      type_location1 = type_location, 
                      type_location = NULL)) %>% 
  map(., ~pivot_wider(.x, names_from = type_location1, values_from = value_1,
                      names_prefix = "V_1_")) %>% 
  map(., ~pivot_wider(.x, names_from = type_location2, values_from = value_2,
                      names_prefix = "V_2_")) %>% 
  map(., ~group_by(.x, ID, age) %>% summarise_all(., na.omit)) %>% 
  bind_rows()

# # A tibble: 8 x 8
# # Groups:   ID [8]
#      ID   age V_1_country V_1_province V_1_city V_2_country V_2_province V_2_city
#   <int> <int> <chr>       <chr>        <chr>    <chr>       <chr>        <chr>   
# 1     1    83 X           A            J        X           A            I       
# 2     2    15 X           NA           K        X           NA           J       
# 3     3     2 Y           B            I        Y           B            I       
# 4     4    12 X           NA           L        Z           NA           K       
# 5     5     2 Y           NA           J        X           NA           J       
# 6     6     2 Y           A            M        Y           B            I       
# 7     7    18 X           B            J        X           A            L       
# 8     8    85 X           A            I        Y           B            M       

数据：

df1 <- read.table(text="ID  age type_location          value_1  value_2    
                        1   83  country:province:city  X:A:J    X:A:I
                        2   15  country:city           X:K      X:J
                        3   2   country:province:city  Y:B:I    Y:B:I
                        4   12  country:city           X:L      Z:K
                        5   2   country:city           Y:J      X:J
                        6   2   country:province:city  Y:A:M    Y:B:I
                        7   18  country:province:city  X:B:J    X:A:L
                        8   85  country:province:city  X:A:I    Y:B:M", 
                  header=T, stringsAsFactors=F)