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我有一组x由 12 列和 167 行组成的数据。第一列是每一行的化合物 ID。我想将t.test3 列作为一组运行，将其他 3 组作为第二组运行，分别针对每一行。我的代码如下，但它不起作用。

for (i in 1:nrow(x)) {
 function(i)c(compound=i,
    t.test(x[2:4],x[8:10],
      x[x$compound==i, ],
      alternative='two.sided',conf.level=0.95)
    )
}
print(c(compound=i,t.test(x[2:4],x[8:10],x[x$compound==i,],
    alternative='two.sided',conf.level=0.95)))

我的意图是t.test对 AC 组和 SC 组之间的每种代谢物（化合物）进行分析，这是两组细胞。

compound    AC-1     AC-2     AC-3     AM-1      AM-2      AM-3      SC-1     SC-2     SC-3     SM-1      SM-2      SM-3
alanine     27612820 22338050 15359640 19741350  18726880  18510800  10914980 12071660 16036180 16890860  16066960  16364300
arginine    7067206  7172234  5933320  136272600 131596800 134717600 6102838  7186256  6770344  140127100 155341300 151748000
asparagine  3151398  2141378  1240904  11522180  8907711   9842342   1677299  2265826  2942991  11690360  12552660  12102620
4

4 回答 4

2

t.test 用于比较两个数据集。可以像这样从矩阵的三个不同列中收集两个数据集：

data_a = c(x[,2:4])
data_b = c(x[,4:8])

此时可以使用 t.test 评估这两个数据集：

t.test(data_a, data_b)

从给定行（氨基酸）的两种不同化合物的三列中收集数据，我们修改并添加一个循环：

x <- matrix(rnorm(24, mean=0, sd=1), 4, ncol=6)
x
           [,1]       [,2]        [,3]      [,4]       [,5]       [,6]
[1,] -0.4810307  0.3996071  0.90663635 0.7487048  0.5787846  2.0231681
[2,] -2.0454921 -0.1225105 -1.04447522 0.9325333 -1.7782776  0.6856150
[3,] -0.3099937  1.2079548 -0.03835271 0.2751349  1.0111554 -0.4862846
[4,] -0.2834953  0.1930481 -0.57968344 0.1204925 -0.5015843  0.3690397

for(i in 1:nrow(x)){
data_a = c(x[i, 1:3])
data_b = c(x[i, 4:6])
print(t.test(data_a, data_b))
}
于 2015-01-23T23:04:14.753 回答
2 
x$stat <- sapply(1:nrow(x), function(i) t.test(as.numeric(as.character(unlist(x[i,2:4]))), as.numeric(as.character(unlist(x[i,8:10]))))[c("p.value")])
于 2017-07-11T13:00:11.280 回答
1

有了这个假数据：

df <- data.frame(compound = c("alanine ", "arginine", "asparagine", "aspartate"))
df <- matrix(rnorm(12*4), ncol = 12)
colnames(df) <- c("AC-1", "AC-2", "AC-3", "AM-1", "AM-2", "AM-3", "SC-1", "SC-2", "SC-3", "SM-1", "SM-2", "SM-3")
df <- data.frame(compound = c("alanine ", "arginine", "asparagine", "aspartate"), df)
 df
    compound        AC.1        AC.2       AC.3       AM.1       AM.2        AM.3       SC.1       SC.2       SC.3         SM.1
1   alanine   1.18362683 -2.03779314 -0.7217692 -1.7569264 -0.8381042  0.06866567  0.2327702 -1.1558879  1.2077454  0.437707310
2   arginine -0.19610110  0.05361113  0.6478384 -0.1768597  0.5905398 -0.67945600 -0.2221109  1.4032349  0.2387620  0.598236199
3 asparagine  0.02540509  0.47880021 -0.1395198  0.8394257  1.9046667  0.31175358 -0.5626059  0.3596091 -1.0963363 -1.004673116
4  aspartate -1.36397906  0.91380826  2.0630076 -0.6817453 -0.2713498 -2.01074098  1.4619707 -0.7257269  0.2851122 -0.007027878
         SM.2        SM.3
1 -0.08419146  0.14275728
2 -1.44965718 -0.64314509
3  0.37673942 -0.07245741
4  0.52794136  1.62305413

您可以执行以下操作来提取（例如）p 值：

library(zoo)
rollapply(t(df[, -1]), function(x) t.test(x)$p.value, width = 3, by = 3)
          [,1]      [,2]      [,3]      [,4]
[1,] 0.6308340 0.5702970 0.5783582 0.6468241
[2,] 0.2511564 0.8327439 0.1617192 0.2005518
[3,] 0.9026407 0.4309623 0.4156030 0.6441069
[4,] 0.3878145 0.4909217 0.6239915 0.2747601
于 2015-01-23T23:22:13.723 回答
1

另一种选择是使用包。

您的数据：

df <- rbind(c(27612820, 22338050, 15359640, 19741350, 18726880, 18510800, 10914980, 12071660, 16036180, 16890860, 16066960, 16364300),
        c(7067206, 7172234, 5933320, 136272600, 131596800, 134717600, 6102838, 7186256, 6770344, 140127100, 155341300, 151748000),
        c(3151398, 2141378, 1240904, 11522180, 8907711, 9842342, 1677299, 2265826, 2942991, 11690360, 12552660, 12102620)
        )

df <- data.frame(df)
rownames(df) <- c("alanine", "arginine", "asparagine")
colnames(df) <- c("AC-1", "AC-2", "AC-3", "AM-1", "AM-2", "AM-3", "SC-1", "SC-2", "SC-3", "SM-1", "SM-2", "SM-3")

然后在 AC 和 SC 组之间的每一行上运行 t 检验：

library(matrixTests)
> row_t_welch(df[,c("AC-1", "AC-2", "AC-3")], df[,c("SC-1", "SC-2", "SC-3")])
           obs.x obs.y obs.tot   mean.x   mean.y mean.diff          var.x         var.y    stderr       df   statistic    pvalue conf.low conf.high alternative mean.null conf.level
alanine        3     3       6 21770170 13007607 8762563.3 37776970798900 7213669482133 3872580.5 2.736945  2.26271945 0.1171389 -4259692  21784819   two.sided         0       0.95
arginine       3     3       6  6724253  6686479   37774.0   471939373529  298723602417  506840.9 3.807645  0.07452832 0.9443398 -1397926   1473474   two.sided         0       0.95
asparagine     3     3       6  2177893  2295372 -117478.7   913496858185  401148784303  661978.3 3.472571 -0.17746605 0.8690016 -2070931   1835973   two.sided         0       0.95
于 2018-06-08T13:01:56.687 回答