R：使用gather和spread将时间序列数据集从宽转长答案

【问题标题】：R: Use gather and spread to transpose a time series dataset from wide to longR：使用gather和spread将时间序列数据集从宽转长
【发布时间】：2020-02-18 00:12:06
【问题描述】：

所以我有一个时间序列数据集，它的形状很宽，不适合分析。我打算做的是使用收集和传播以我想要的方式将数据转换为长。

样本数据：

df <- data.frame(
  id = 1:3,
  "X1" = c(100,90,80),
  "X2" = c(100,90,80),
  "X3" = c(100,90,80),
  "X4" = c(100,90,80),
  "X5" = c(100,90,80)
)

> print(df)
  id  X1  X2  X3  X4  X5
1  1 100 100 100 100 100
2  2  90  90  90  90  90
3  3  80  80  80  80  80

期望的输出：

df2 <- data.frame(
  id = c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3),
  "month" = c(1,2,3,4,5,1,2,3,4,5,1,2,3,4,5),
  "balance" = c(100,100,100,100,100,90,90,90,90,90,80,80,80,80,80)
)
print(df2)

> print(df2)
   id month balance
1   1     1     100
2   1     2     100
3   1     3     100
4   1     4     100
5   1     5     100
6   2     1      90
7   2     2      90
8   2     3      90
9   2     4      90
10  2     5      90
11  3     1      80
12  3     2      80
13  3     3      80
14  3     4      80
15  3     5      80

我尝试了什么，我得到了什么：

test_gather <-gather(df, "month", "value", X1:X5,na.rm = FALSE )

> print(test_gather)
   id month value
1   1    X1   100
2   2    X1    90
3   3    X1    80
4   1    X2   100
5   2    X2    90
6   3    X2    80
7   1    X3   100
8   2    X3    90
9   3    X3    80
10  1    X4   100
11  2    X4    90
12  3    X4    80
13  1    X5   100
14  2    X5    90
15  3    X5    80

如上所示，我得到的结果根本没有正确排序。如果可能的话，我希望能够直接通过收集/传播而不在 dplyr 中进行安排来获得所需的输出。

【问题讨论】：

标签： r

【解决方案1】：

这需要在gather 之后的“id”列上添加arrange 并删除前缀（如有必要）

library(stringr)
library(dplyr)
library(tidyr)
df %>%
   gather(month, balance, -id) %>%
   arrange(id) %>%
   mutate(month = str_remove(month, "^X"))
#   id month balance
#1   1     1     100
#2   1     2     100
#3   1     3     100
#4   1     4     100
#5   1     5     100
#6   2     1      90
#7   2     2      90
#8   2     3      90
#9   2     4      90
#10  2     5      90
#11  3     1      80
#12  3     2      80
#13  3     3      80
#14  3     4      80
#15  3     5      80

使用来自tidyr_1.0.0 的pivot_longer，它不会改变顺序

df %>%
   pivot_longer(cols = -id, names_to = "month", values_to = "balance") %>%
   mutate(month = readr::parse_number(month))
# A tibble: 15 x 3
#      id month balance
#   <int> <dbl>   <dbl>
# 1     1     1     100
# 2     1     2     100
# 3     1     3     100
# 4     1     4     100
# 5     1     5     100
# 6     2     1      90
# 7     2     2      90
# 8     2     3      90
# 9     2     4      90
#10     2     5      90
#11     3     1      80
#12     3     2      80
#13     3     3      80
#14     3     4      80
#15     3     5      80

【讨论】：

【解决方案2】：

基础解决方案：

# Reshape from wide to long: 

df2 <- 

  data.frame(

    reshape(

      df,

      direction = "long",

      varying = names(df)[names(df) != "id"],

      v.names = "value",

      idvar = "id",

      timevar = "month",

      times = names(df)[names(df) != "id"]

    ),

  row.names = NULL

)

# Order by id:

df2 <- df2[order(df2$id),]

# Keep only month numbers and coerce vector to numeric: 

df2$month <- as.numeric(gsub("^\\D+", "", df2$month))

数据

df <- data.frame(

  id = 1:3,

  "X1" = c(100,90,80),

  "X2" = c(100,90,80),

  "X3" = c(100,90,80),

  "X4" = c(100,90,80),

  "X5" = c(100,90,80)
)

【讨论】：