Question

我希望将660列的数据帧转换为3列，只需将它们按照每3列相互堆叠，而无需手动重新排列（因为我有660列）。

在每2列只有4列的小规模示例中，我想从

开始

到

iota

由于

Answer 1

reshape救援：

reshape(df, direction="long", varying=split(names(df), rep(seq_len(ncol(df)/2), 2)))

#    time A  B id
#1.1    1 1  4  1
#2.1    1 2  5  2
#3.1    1 3  6  3
#1.2    2 7 10  1
#2.2    2 8 11  2
#3.2    2 9 12  3

Answer 2

经典的分组 - 应用 - 组合方法将灵活扩展：

as.data.frame(lapply(split(unclass(df), 
                           names(df)[seq(ncol(df) / 2)]), 
                     unlist, use.names = FALSE))
##   A  B
## 1 1  4
## 2 2  5
## 3 3  6
## 4 7 10
## 5 8 11
## 6 9 12

或提示purrr，

library(purrr)

df %>% unclass() %>%    # convert to non-data.frame list
    split(names(.)[seq(length(.) / 2)]) %>%     # split columns by indexed names
    map_df(simplify)    # simplify each split to vector, coerce back to data.frame
## # A tibble: 6 × 2
##       A     B
##   <int> <int>
## 1     1     4
## 2     2     5
## 3     3     6
## 4     7    10
## 5     8    11
## 6     9    12

Answer 3

rbind.data.frame要求所有列都匹配。因此，使用setNames替换C：D列的名称：

rbind( dat[1:2], setNames(dat[3:4], names(dat[1:2]))  )
  A  B
1 1  4
2 2  5
3 3  6
4 7 10
5 8 11
6 9 12

要将其概括为多列，请使用do.call和lapply：

 dat <- setNames( as.data.frame( matrix(1:36, ncol=12) ), LETTERS[1:12])
 dat
#----
  A B C  D  E  F  G  H  I  J  K  L
1 1 4 7 10 13 16 19 22 25 28 31 34
2 2 5 8 11 14 17 20 23 26 29 32 35
3 3 6 9 12 15 18 21 24 27 30 33 36


do.call( rbind, lapply( seq(1,12, by=3), function(x) setNames(dat[x:(x+2)], LETTERS[1:3])   ))
    A  B  C
1   1  4  7
2   2  5  8
3   3  6  9
4  10 13 16
5  11 14 17
6  12 15 18
7  19 22 25
8  20 23 26
9  21 24 27
10 28 31 34
11 29 32 35
12 30 33 36

12将被660取代，其他一切都应该有效。

Answer 4

这是另一个let headers = new Headers(); headers.append('Content-Type', 'application/json'); headers.append('authentication', `${student.token}`); let options = new RequestOptions({ headers: headers }); return this.http .put(url, JSON.stringify(student), options)选项

base R

或其他选项i1 <- c(TRUE, FALSE) `row.names<-`(data.frame(A= unlist(df1[i1]), B = unlist(df1[!i1])), NULL) # A B #1 1 4 #2 2 5 #3 3 6 #4 7 10 #5 8 11 #6 9 12来自melt

data.table

Answer 5

来自rbindlist包的

data.table也可以用于该任务，并且似乎更有效。

# EXAMPLE DATA 
df1 <- read.table(text = '
Col1 Col2 Col3 Col4
1    2    3    4
5    6    7    8
1    2    3    4
5    6    7    8', header = TRUE)

library(data.table)
library(microbenchmark)
library(purrr)
microbenchmark(
    Map = as.data.frame(Map(c, df1[,1:2], df1[, 3:4])),
    Reshape = reshape(df1, direction="long", varying=split(names(df1), rep(seq_len(ncol(df1)/2), 2))),
    Purr = df1 %>% unclass() %>%    # convert to non-data.frame list
        split(names(.)[seq(length(.) / 2)]) %>%     # split columns by indexed names
        map_df(simplify), 
    DataTable = rbindlist(list(df1[,1:2], df1[, 3:4])),
    Mapply = data.frame(mapply(c, df1[,1:2], df1[, 3:4], SIMPLIFY=FALSE)),
    Rbind = rbind(df1[, 1:2],setnames(df1[, 3:4],names(df1[,1:2])))
)

结果是：

Unit: microseconds
expr     min         lq       mean         median     uq         max neval      cld
Map        214.724   232.9380  246.2936    244.1240   255.9240   343.611    100  bc  
Reshape    716.962   739.8940  778.7912    749.7550   767.6725   2834.192   100     e
Purr       309.559   324.6545  339.2973    334.0440   343.4290   551.746    100    d 
DataTable  98.228    111.6080  122.7753    119.2320   129.2640   189.614    100 a    
Mapply     233.577   258.2605  271.1881    270.7895   281.6305   339.291    100   c  
Rbind      206.001   221.1515  228.5956    226.6850   235.2670   283.957    100  b

如何在R中按x列堆叠列

5 个答案: