Question

我有一个df：

   group number id
1      A   abcd  1
2      A   abcd  2
3      A   abcd  3
4      A   efgh  4
5      A   efgh  5
6      B   abcd  1
7      B   abcd  2
8      B   abcd  3
9      B   abcd  9
10     B   ijkl 10

我想这样做：

   group number  data1 data2 data3 data4           Length
1      A   abcd      1     2     3                      3
2      A   efgh      4     5                            2
3      B   abcd      1     2     3     9                4
4      B   ijkl      10                                 1

对不起，我只能这样做到df2：

   group number     data               Length
1      A   abcd  c(1,2,3)                   3
2      A   efgh  c(4,5)                     2
3      B   abcd  c(1,2,3,9)                 4
4      B   ijkl  10                         1

我的代码在这里：

library(tidyverse)

df <- data.frame (group = c(rep('A',5),rep("B",5)),
                  number = c(rep('abcd',3),rep('efgh',2),rep('abcd',4),rep('ijkl',1)),
                  id = c(1,2,3,4,5,1,2,3,9,10))

df2 <- df %>%
  group_by(group,number) %>%
  nest() %>%
  mutate(data=map(data,~unlist(.x, recursive = TRUE, use.names = FALSE))，
         Length= map(data, ~length(.x)))

请随意从df或df2开始，（out）任何包都没问题。

Answer 1

您可以将名称count更改为length（另外，我将{＆＃39;空格＆＃39;更改为NA，如果要更改，{{1 }}）

选项1

df2[is.na(df2)]=''

选项2

df <- data.frame (group = c(rep('A',5),rep("B",5)),
                  number = c(rep('abcd',3),rep('efgh',2),rep('abcd',4),rep('ijkl',1)),
                  id = c(1,2,3,4,5,1,2,3,9,10))

df2 <- df %>%
    group_by(group,number) %>%
    mutate(data=toString(id),count=n())

library(splitstackshape)
cSplit(df2, 3, drop = TRUE,sep=',')


   group number count data_1 data_2 data_3 data_4
1:     A   abcd     3      1      2      3     NA
2:     A   efgh     2      4      5     NA     NA
3:     B   abcd     4      1      2      3      9
4:     B   ijkl     1     10     NA     NA     NA

Answer 2

我必须盲目地把它给你，但这应该有效或接近：

library(tidyverse)
df %>%
    group_by(group,number) %>%
    mutate(key = paste0("data",row_number()),length = n()) %>%
    ungroup %>%
    spread(key,id,"")

为了使它能够从你的嵌套数据中运行，我认为你必须将这些向量更改为1行data.frames具有相同的col号和名称，然后使用unexst，更复杂！：）

Answer 3

在基地R

temp = split(df, paste(df$group, df$number))
columns = max(sapply(temp, NROW))
do.call(rbind, lapply(temp, function(a)
    cbind(group = a$group[1],
          number = a$number[1],
          setNames(data.frame(t(a$id[1:columns])), paste0("data", 1:columns)),
          length = length(a$id))
))
#       group number data1 data2 data3 data4 length
#A abcd     A   abcd     1     2     3    NA      3
#A efgh     A   efgh     4     5    NA    NA      2
#B abcd     B   abcd     1     2     3     9      4
#B ijkl     B   ijkl    10    NA    NA    NA      1

Answer 4

以下是使用data.table

的选项

library(data.table)
dcast(setDT(df), group + number~ paste0("data", rowid(group, number)), 
 value.var = 'id', fill = 0)[, 
   length := Reduce(`+`, lapply(.SD, `>`, 0)), .SDcols = data1:data4][]
#    group number data1 data2 data3 data4 length
#1:     A   abcd     1     2     3     0      3
#2:     A   efgh     4     5     0     0      2
#3:     B   abcd     1     2     3     9      4
#4:     B   ijkl    10     0     0     0      1

Answer 5

这是[{3}}的变体，它在从长格式转换为宽格式之前计算Length ，并在调用{prefix时使用rowid()参数1}}：

library(data.table) data.table(df)[, Length := .N, by = .(group, number)][ , dcast(.SD, group + number + Length ~ rowid(group, number, prefix = "data"), value.var = "id")]

group number Length data1 data2 data3 data4 1: A abcd 3 1 2 3 NA 2: A efgh 2 4 5 NA NA 3: B abcd 4 1 2 3 9 4: B ijkl 1 10 NA NA NA

对于漂亮的打印，NA值可以转换为空格：

data.table(df)[, Length := .N, by = .(group, number)][ , dcast(.SD, group + number + Length ~ rowid(group, number, prefix = "data"), as.character, value.var = "id", fill = "")]

group number Length data1 data2 data3 data4 1: A abcd 3 1 2 3 2: A efgh 2 4 5 3: B abcd 4 1 2 3 9 4: B ijkl 1 10

传染媒介的元素对数据框架的不同的专栏

5 个答案: