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
我需要重塑一下这样的事情：
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
我知道我应该使用融化，但我不知道该怎么做。
我需要为互联网神灵牺牲一些单词/字符让我发布这个问题;）.................
答案 0 :(得分：3)
我认为这应该是正确的方式。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
答案 1 :(得分：1)
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 ]