这就是我的数据的样子：
> dput(data)
structure(list(Name = c("Mark", "Tere", "Marcus", "Heidi", "Georg",
"Tieme", "Joan", "Tytus", "Mark", "Tere", "Marcus", "Heidi",
"Georg", "Tieme", "Joan", "Tytus", "Mark", "Tere", "Marcus",
"Heidi", "Georg", "Tieme", "Joan", "Tytus", "Mark", "Tere", "Marcus",
"Heidi", "Georg", "Tieme", "Joan", "Tytus", "Mark", "Tere", "Marcus",
"Heidi", "Georg", "Tieme", "Joan", "Tytus", "Mark", "Tere", "Marcus",
"Heidi", "Georg", "Tieme", "Joan", "Tytus"), position = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("n",
"w"), class = "factor"), time = c(0, 0, 0, 0, 0, 0, 0, 0, 1,
1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 0, 0, 0, 0, 0, 0,
0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2), type = c(5,
5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5,
5, 5, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3), value = structure(c(0.314543267094949, 3.19693140536703,
0.930679293700748, 0.344465436879545, 1.12925174828143, 0.295226529240608,
0.534869402647018, 0.699742971466516, 0.899450918860775, 1.69467910557034,
0.105376912280917, 2.49469271475885, 0.0343013256788254, 0.469440043556817,
0.917351596015406, 1.46852763733807, 0.967872510622156, 0.354032471310347,
1.52519534333589, 1.04999953913746, 1.90298544048312, 0.265492846723646,
1.19947244036467, 2.54636084834549, 0.401142360642552, 0.190538682818794,
1.55882554492893, 1.27338900133492, 0.298795571085066, 1.73941845252159,
0.133721855003387, 0.115250693634152, 1.12742130434271, 1.50438847943189,
0.0286978152580559, 0.740798383999788, 0.838944700504553, 0.726393882538543,
1.79097043714466, 0.214124974329025, 1.46675111903789, 0.0172332774632657,
2.89433199895506, 3.07798819227104, 0.826817313167167, 0.72394085004025,
0.083957624156028, 1.07571752737732), .Dim = c(48L, 1L))), .Names = c("Name",
"position", "time", "type", "value"), row.names = c(NA, 48L), class = "data.frame")
我想对value特定行执行t检验。如何找到要执行t检验的行：
1. Name必须相同。
2. Type必须相同。
3. Time =0总是那么t检验之间应计算出的参考time=1和time= 0，time=2和time= 0，等。这是一个示例数据，因此我无法告诉您time实际数据中有多少，但0始终是参考。
4.输出（t检验的结果）可以存储在指定的附加列和行中time。
5.数的Names和Times我不知道。数据包含数千行。
6.我相信循环是唯一的解决方案，无论如何，这是我想要的代码。当然，没有循环的其他解决方案也非常受欢迎。
这是一些编程，所以我将开始赏金奖励那些愿意为此付出努力的人。
我们可以尝试 data.table
library(data.table)
setDT(data, keep.rownames= TRUE)
dt1 <- data[time==0]
dt2 <- data[time!=0]
res <- dt2[dt1, on = c("Name", "type")][, .(r1= rn, r2 = i.rn, time0 = i.time,
Othertime = time, pval = t.test(value, i.value)$p.value), .(Name, type)]
res1 <- melt(res, measure = patterns("^r\\d+", "time"),
value.name= c("rn", "time"))[, variable := NULL][]
res2 <- unique(res1, by = c("Name", "type", "rn"))
res3 <- data[res2, on = "rn"][order(as.numeric(rn))][,
c('i.Name', 'i.type', 'i.time') := NULL][]
head(res3,3)
# rn Name position time type value pval
#1: 1 Mark n 0 5 0.3145433 0.03514213
#2: 2 Tere n 0 5 3.1969314 0.19052234
#3: 3 Marcus n 0 5 0.9306793 0.89741695
使用OP的真实数据
file <- "C:/Users/akrun/Misc/Data_to_analyze.csv"
data1 <- fread(file)
dt1 <- data1[tp==0]
dt2 <- data1[tp!=0]
res <- dt2[dt1, on = c("protein", "fract"), allow.cartesian = TRUE
][, if(.N >1) {
.(r1= V1, r2 = i.V1, time0 = i.tp,Othertime = tp,
pval = t.test(abundance, i.abundance)$p.value)
} # else .(r1=V1, r2 = i.V1, time0 = i.tp, Othertime = tp, pval = NA_real_)
, by = .(protein, fract)]
res1 <- melt(res, measure = patterns("^r\\d+", "time"),
value.name= c("V1", "time"))[, variable := NULL][]
res2 <- unique(res1, by = c("protein", "fract", "V1"))
res3 <- data1[res2, on = "V1"][order(as.numeric(V1))][,
c('i.protein', 'i.fract') := NULL][]
head(res3)
# V1 protein line tp fract abundance pval time
#1: 1 PCP000002 n 0 1 46697305.3 0.515626208527 1
#2: 2 PCP000007 n 0 1 8384565.8 0.000643157099 1
#3: 3 PCP000008 n 0 1 570026.3 0.000006871077 1
#4: 4 PCP000012 n 0 1 1018257.3 0.085610886030 1
#5: 5 PCP000017 n 0 1 157877521.4 0.016528856828 1
#6: 6 PCP000018 n 0 1 426215586.5 0.046130079694 1
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我来说两句