R-将数据两列重塑为1答案

【问题标题】：R- Reshaping Data two columns to 1R-将数据两列重塑为1
【发布时间】：2017-12-31 06:43:22
【问题描述】：

您好，我有以下数据框：

        Date          S1    Mavg1          STdev1         S2     Mavg2        STdev2
1 02/12/1999 -0.000217052  0.002862195 -0.002390842  2.2025825 -1.17258213 -0.3057015
2 03/12/1999 -0.004882038 -0.015920939 -0.007014382 -0.4040079 -0.69496488  0.1000167
3 06/12/1999 -0.001445954  0.005077610  0.000000000  1.1573779 -1.64268166 -0.3847015
4 07/12/1999 -0.000907952 -0.005403168  0.002027728 -0.3165827 -1.16466940 -1.0113501
5 08/12/1999  0.000881220  0.011375226 -0.013524439  0.5154455 -0.51725208  1.0942177
6 09/12/1999  0.001043752  0.013400502  0.017033342  1.9368608 -0.05587143  0.5203337

我需要重新塑造它以拥有这样的东西：

         Date  S variable        value
1  02/12/1999 S1    value -0.000217052
2  03/12/1999 S1    value -0.004882038
3  06/12/1999 S1    value -0.001445954
4  07/12/1999 S1    value -0.000907952
5  08/12/1999 S1    value  0.000881220
6  09/12/1999 S1    value  0.001043752
7  02/12/1999 S2    value  2.202582500
8  03/12/1999 S2    value -0.404007900
9  06/12/1999 S2    value  1.157377900
10 07/12/1999 S2    value -0.316582700
11 08/12/1999 S2    value  0.515445500
12 09/12/1999 S2    value  1.936860800
13 02/12/1999 S1     mavg  0.002862195
14 03/12/1999 S1     mavg -0.015920939
15 06/12/1999 S1     mavg  0.005077610
16 07/12/1999 S1     mavg -0.005403168
17 08/12/1999 S1     mavg  0.011375226
18 09/12/1999 S1     mavg  0.013400502
19 02/12/1999 S2     mavg -1.172582130
20 03/12/1999 S2     mavg -0.694964880
21 06/12/1999 S2     mavg -1.642681660
22 07/12/1999 S2     mavg -1.164669400
23 08/12/1999 S2     mavg -0.517252080
24 09/12/1999 S2     mavg -0.055871430
25 02/12/1999 S1   stddev -0.002390842
26 03/12/1999 S1   stddev -0.007014382
27 06/12/1999 S1   stddev  0.000000000
28 07/12/1999 S1   stddev  0.002027728
29 08/12/1999 S1   stddev -0.013524439
30 09/12/1999 S1   stddev  0.017033342
31 02/12/1999 S2   stddev -0.305701500
32 03/12/1999 S2   stddev  0.100016700
33 06/12/1999 S2   stddev -0.384701500
34 07/12/1999 S2   stddev -1.011350100
35 08/12/1999 S2   stddev  1.094217700
36 09/12/1999 S2   stddev  0.520333700

我知道我应该使用 melt，但我不知道该怎么做。

我需要牺牲一些文字/字符让互联网大神让我发布这个问题;) ..................

【问题讨论】：

到目前为止你尝试过什么？您可以在easy-to-paste form 中提供您的示例吗？
这是一个易于粘贴的表单，是的，我确实尝试过，但它不起作用

标签： r reshape2 melt

【解决方案1】：

我认为这应该是正确的方法。我使用了tidyverse 方法：

mydf %>% 
    gather(key=Variable,value=Value,c(S1, S2, Mavg1, Mavg2,STdev1,STdev2)) %>% 
    mutate(S=case_when(
        Variable=="S1"|Variable=="Mavg1"|Variable=="STdev1" ~ "S1",
        Variable=="S2"|Variable=="Mavg2"|Variable=="STdev2" ~ "S2"
    ),
    Variable=str_replace_all(string=Variable,pattern="S[1|2]",replacement="Value"),
    Variable=str_replace_all(string=Variable,pattern="[[:digit:]]",replacement="")
    ) %>% 
    select(Date,S,Variable,Value)

这是输出：

         Date  S Variable        Value
1  02/12/1999 S1    Value -0.000217052
2  03/12/1999 S1    Value -0.004882038
3  06/12/1999 S1    Value -0.001445954
4  07/12/1999 S1    Value -0.000907952
5  08/12/1999 S1    Value  0.000881220
6  09/12/1999 S1    Value  0.001043752
7  02/12/1999 S2    Value  2.202582500
8  03/12/1999 S2    Value -0.404007900
9  06/12/1999 S2    Value  1.157377900
10 07/12/1999 S2    Value -0.316582700
11 08/12/1999 S2    Value  0.515445500
12 09/12/1999 S2    Value  1.936860800
13 02/12/1999 S1     Mavg  0.002862195
14 03/12/1999 S1     Mavg -0.015920939
15 06/12/1999 S1     Mavg  0.005077610
16 07/12/1999 S1     Mavg -0.005403168
17 08/12/1999 S1     Mavg  0.011375226
18 09/12/1999 S1     Mavg  0.013400502
19 02/12/1999 S2     Mavg -1.172582130
20 03/12/1999 S2     Mavg -0.694964880
21 06/12/1999 S2     Mavg -1.642681660
22 07/12/1999 S2     Mavg -1.164669400
23 08/12/1999 S2     Mavg -0.517252080
24 09/12/1999 S2     Mavg -0.055871430
25 02/12/1999 S1    STdev -0.002390842
26 03/12/1999 S1    STdev -0.007014382
27 06/12/1999 S1    STdev  0.000000000
28 07/12/1999 S1    STdev  0.002027728
29 08/12/1999 S1    STdev -0.013524439
30 09/12/1999 S1    STdev  0.017033342
31 02/12/1999 S2    STdev -0.305701500
32 03/12/1999 S2    STdev  0.100016700
33 06/12/1999 S2    STdev -0.384701500
34 07/12/1999 S2    STdev -1.011350100
35 08/12/1999 S2    STdev  1.094217700
36 09/12/1999 S2    STdev  0.520333700

【讨论】：

它回答了这个问题吗？查看输出中的行数。它应该是 36 行而不是 60 行。查看我的解决方案，输出中有 36 行
@Sathish 我认为你绝对是对的，我已经重新考虑了我的代码。现在输出是正确的。非常感谢

【解决方案2】：

library('data.table') 
df2 <- melt(df1, id.vars = c("Date"))
# assign values to column "S" by finding the numbers at the end of values in variable column
# and paste it with "S"
df2[, S := paste0("S", unlist(lapply( regmatches(variable, gregexpr('[0-9]+', variable )), tail, 1 ))) ]
# find and replace S1 and S2 with "value" in variable column
df2[variable %in% c("S1", "S2"), variable := "value"]
df2[, variable := gsub("[0-9]", "", variable)]
df2
#          Date variable        value  S
# 1: 02/12/1999    value -0.000217052 S1
# 2: 03/12/1999    value -0.004882038 S1
# 3: 06/12/1999    value -0.001445954 S1
# 4: 07/12/1999    value -0.000907952 S1
# 5: 08/12/1999    value  0.000881220 S1
# 6: 09/12/1999    value  0.001043752 S1
# 7: 02/12/1999     Mavg  0.002862195 S1
# 8: 03/12/1999     Mavg -0.015920939 S1
# 9: 06/12/1999     Mavg  0.005077610 S1
# 10: 07/12/1999     Mavg -0.005403168 S1
# 11: 08/12/1999     Mavg  0.011375226 S1
# 12: 09/12/1999     Mavg  0.013400502 S1
# 13: 02/12/1999    STdev -0.002390842 S1
# 14: 03/12/1999    STdev -0.007014382 S1
# 15: 06/12/1999    STdev  0.000000000 S1
# 16: 07/12/1999    STdev  0.002027728 S1
# 17: 08/12/1999    STdev -0.013524439 S1
# 18: 09/12/1999    STdev  0.017033342 S1
# 19: 02/12/1999    value  2.202582500 S2
# 20: 03/12/1999    value -0.404007900 S2
# 21: 06/12/1999    value  1.157377900 S2
# 22: 07/12/1999    value -0.316582700 S2
# 23: 08/12/1999    value  0.515445500 S2
# 24: 09/12/1999    value  1.936860800 S2
# 25: 02/12/1999     Mavg -1.172582130 S2
# 26: 03/12/1999     Mavg -0.694964880 S2
# 27: 06/12/1999     Mavg -1.642681660 S2
# 28: 07/12/1999     Mavg -1.164669400 S2
# 29: 08/12/1999     Mavg -0.517252080 S2
# 30: 09/12/1999     Mavg -0.055871430 S2
# 31: 02/12/1999    STdev -0.305701500 S2
# 32: 03/12/1999    STdev  0.100016700 S2
# 33: 06/12/1999    STdev -0.384701500 S2
# 34: 07/12/1999    STdev -1.011350100 S2
# 35: 08/12/1999    STdev  1.094217700 S2
# 36: 09/12/1999    STdev  0.520333700 S2
#           Date variable        value  S

数据：

library('data.table') 
df1 <- fread('id       Date          S1    Mavg1          STdev1         S2     Mavg2        STdev2
1 02/12/1999 -0.000217052  0.002862195 -0.002390842  2.2025825 -1.17258213 -0.3057015
      2 03/12/1999 -0.004882038 -0.015920939 -0.007014382 -0.4040079 -0.69496488  0.1000167
      3 06/12/1999 -0.001445954  0.005077610  0.000000000  1.1573779 -1.64268166 -0.3847015
      4 07/12/1999 -0.000907952 -0.005403168  0.002027728 -0.3165827 -1.16466940 -1.0113501
      5 08/12/1999  0.000881220  0.011375226 -0.013524439  0.5154455 -0.51725208  1.0942177
      6 09/12/1999  0.001043752  0.013400502  0.017033342  1.9368608 -0.05587143  0.5203337', header = TRUE)
df1[, id := NULL ]

【讨论】：

在任何情况下，您的输出都包含STdev1 或STdev2 之类的值。 OP 想要在 variable 值中没有数字的输出；-)
一个gsub 会处理它df2[, variable := gsub("[0-9]", "", variable)]