我正在尝试自动化下面的R代码，其中我正在计算p值。数据为csv格式。 我有每个部分及其版本的点击次数和公开号码。 如果有人可以帮助您应用任何循环或其他内容。
我具有.csv格式的数据：
` Total Clicks
Section Version A Version B Version C Version D
Section1 1,999 2,116 2,307 2,568
Section2 3,450 1,781 3,416 1,399
Section3 1,773 915 1,744 644
Section4 0 2,255 0 1,432
Section5 588 573 721 235
Main email 7,222 7,067 7,467 6,043
Total email 7,810 7,640 8,188 6,278
`
`Version # Opens
A 9,073
B 9,150
C 9,215
D 9,153
`
目前，我正在以以下格式手动分配数据：
` S1_Click_A=1,999 ####(section 1, email A)
S1_Click_B=2,116 ## (section 1, email B)
S1_Click_C=2,307
S1_Click_D=2,568
S2_Click_A=3,450
S2_Click_B=1,781
.
.
.
S5_Click_C=721
S5_Click_D=235
MainBody_Click_A=7,222
MainBody_Click_B=7,067
.
.
TotalEmail_Click_C=8,188
TotalEmail_Click_D=6,278
`
`# to test % total click is the comparable across versions`
`# section 1 test
S1ab <- prop.test(x = c(S1_Click_A,S1_Click_B), n = c(Open_A,Open_B))
S1ac <- prop.test(x = c(S1_Click_A,S1_Click_C), n = c(Open_A,Open_C))
S1ad <- prop.test(x = c(S1_Click_A,S1_Click_D), n = c(Open_A,Open_D))
S1bc <- prop.test(x = c(S1_Click_B,S1_Click_C), n = c(Open_B,Open_C))
S1bd <- prop.test(x = c(S1_Click_B,S1_Click_D), n = c(Open_B,Open_D))
S1cd <- prop.test(x = c(S1_Click_C,S1_Click_D), n = c(Open_C,Open_D))
`
`#section 2 test
S2ab <- prop.test(x = c(S2_Click_A,S2_Click_B), n = c(Open_A,Open_B))
S2ac <- prop.test(x = c(S2_Click_A,S2_Click_C), n = c(Open_A,Open_C))
```
S2cd <- prop.test(x = c(S2_Click_C,S2_Click_D), n = c(Open_C,Open_D))
`
`#section 3 test
S3ab <- prop.test(x = c(S3_Click_A,S3_Click_B), n = c(Open_A,Open_B))
S3ac <- prop.test(x = c(S3_Click_A,S3_Click_C), n = c(Open_A,Open_C))
```
S3cd <- prop.test(x = c(S3_Click_C,S3_Click_D), n = c(Open_C,Open_D))`
`#section 4 test
S4ab <- prop.test(x = c(S4_Click_A,S4_Click_B), n = c(Open_A,Open_B))
S4ac <- prop.test(x = c(S4_Click_A,S4_Click_C), n = c(Open_A,Open_C))
`#section 5 test
S5ab <- prop.test(x = c(S5_Click_A,S5_Click_B), n = c(Open_A,Open_B))
S5ac <- prop.test(x = c(S5_Click_A,S5_Click_C), n = c(Open_A,Open_C))
`#Main body test
MainBodyab <- prop.test(x = c(MainBody_Click_A,MainBody_Click_B), n = c(Open_A,Open_B))
MainBodyac <- prop.test(x = c(MainBody_Click_A,MainBody_Click_C), n = c(Open_A,Open_C))
`
```
`
`#FINAL P VALUE`
`S1ab$p.value
S1ac$p.value
S1ad$p.value
`
I expect:
1. I want to read the data in the above format. I mean reading the
section 1 version A data i.e 1,999 and assigning the same to
S1_Click_A=1,999 similarly for others.
2. a matrix with their clicks and p values in a single row.
dput()
structure(list(Section = structure(c(2L, 3L, 4L, 5L, 6L, 1L, 7L), .Label =
c("Main email body", "Section 1", "Section 2", "Section 3", "Section 4",
"Section 5", "Total email"), class = "factor"), Version.A = c(2967L, 4840L,
2508L, 2093L, 1117L, 12408L, 13525L), Version.B = c(3353L, 4522L, 2250L,
1333L, 925L, 11458L, 12383L), Version.C = c(495L, 285L, 228L, 209L, 186L,
282L, 271L), Version.D = c(559L, 266L, 205L, 133L, 154L, 260L, 248L)), class
= "data.frame", row.names = c(NA, -7L ))
最终格式的输入
structure(list(Section = structure(c(2L, 3L, 4L, 5L, 6L, 1L,
7L), .Label = c("Main email body", "Section 1", "Section 2",
"Section 3", "Section 4", "Section 5", "Total email"), class = "factor"),
Version.A = structure(c(3L, 4L, 2L, 1L, 5L, 6L, 7L), .Label = c("0",
"1,773", "1,999", "3,450", "588", "7,222", "7,810"), class = "factor"),
Version.B = structure(c(2L, 1L, 7L, 3L, 4L, 5L, 6L), .Label = c("1,781",
"2,116", "2,255", "573", "7,067", "7,640", "915"), class = "factor"),
Version.C = structure(c(3L, 4L, 2L, 1L, 6L, 5L, 7L), .Label = c("0",
"1,744", "2,307", "3,416", "7,467", "721", "8,188"), class = "factor"),
Version.D = structure(c(3L, 1L, 7L, 2L, 4L, 5L, 6L), .Label = c("1,399",
"1,432", "2,568", "235", "6,043", "6,278", "644"), class = "factor"),
A.vs..B = c(NA, NA, NA, NA, NA, NA, NA), A.vs..C = c(NA,
NA, NA, NA, NA, NA, NA), A.vs..D = c(NA, NA, NA, NA, NA,
NA, NA), B.vs..C = c(NA, NA, NA, NA, NA, NA, NA), B.vs..D = c(NA,
NA, NA, NA, NA, NA, NA), C.vs..D = c(NA, NA, NA, NA, NA,
NA, NA)), class = "data.frame", row.names = c(NA, -7L))
答案 0 :(得分：1)
这里是第一部分的解决方案，其他部分都是相同的原理。
首先生成组合，然后对其进行测试。
df <- structure(list(Section = structure(c(2L, 3L, 4L, 5L, 6L, 1L, 7L), .Label =
c("Main email body", "Section 1", "Section 2", "Section 3", "Section 4",
"Section 5", "Total email"), class = "factor"), Version.A = c(2967L, 4840L,
2508L, 2093L, 1117L, 12408L, 13525L), Version.B = c(3353L, 4522L, 2250L,
1333L, 925L, 11458L, 12383L), Version.C = c(495L, 285L, 228L, 209L, 186L,
282L, 271L), Version.D = c(559L, 266L, 205L, 133L, 154L, 260L, 248L)), class
= "data.frame", row.names = c(NA, -7L ))
opens <- data.frame(A = 9073, B = 9150, C = 9215, D = 9153)
s1_comb <- combn(colnames(df)[-1], 2)
open_comb <- combn(colnames(opens), 2)
res <- cbind(df,
matrix(NA,
nrow = nrow(df),
ncol = ncol(open_comb),
dimnames = list(1:nrow(df), paste(open_comb[1,], "vs", open_comb[2,]))
)
)
for(k in 1:ncol(s1_comb)){
for(o in 1:(nrow(df)-2)){
res[o, paste(open_comb[1, k], "vs", open_comb[2, k])] <- prop.test(x = unlist(df[o, s1_comb[,k]]),
n = unlist(opens[open_comb[,k]]))$p.value
}
}
res
最终输出，根据要求：
> res
Section Version.A Version.B Version.C Version.D A vs B A vs C A vs D B vs C B vs D
1 Section 1 2967 3353 495 559 2.452892e-08 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
2 Section 2 4840 4522 285 266 1.259231e-07 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
3 Section 3 2508 2250 228 205 2.961113e-06 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
4 Section 4 2093 1333 209 133 1.081110e-48 0.000000e+00 0.000000e+00 4.567813e-198 6.505394e-234
5 Section 5 1117 925 186 154 2.756287e-06 7.420214e-161 3.232226e-174 1.051130e-116 4.618885e-129
6 Main email body 12408 11458 282 260 NA NA NA NA NA
7 Total email 13525 12383 271 248 NA NA NA NA NA
C vs D
1 3.472031e-02
2 4.850847e-01
3 3.178608e-01
4 5.557843e-05
5 1.022220e-01
6 NA
7 NA
答案 1 :(得分：0)