1

我有data.frame df1：

df1 <- data.frame(
  En_ID = c("KNT00000000003", "KNT00000000005", "KNT00000000419", 
                 "KNT00000000457", "KNT00000000460", "KNT00000000938", 
                 "KNT00000000971", "KNT00000001036", "KNT00000001084", 
                 "KNT00000001167" ), 
  `Nor1` = c(-0.834165161710272, 1.02199443531549, 
                -0.558658947885705, -0.390114219973209, -1.23551839713296, 
                3.11429434221998, 0.283932163407262, -1.16908518620064, 
                -0.597054772455507, -0.593624543273255), 
  `Nor2` = c(-1.18531035488942, 0.423719727339646, -1.23261719368372, 
                0.0855281133529292, -1.52366830232278, 3.36692586561211, 
                1.00323690950956, -0.000211248816114964, -4.74738483548391, 
                -0.318176231083024), 
  `Nor3` = c(-0.262659255267546, 1.3962481061442, -0.548673555705647, 
                -0.0149651083306594, -1.45458689193089, 2.54126941463459, 
                1.17711308509307, -1.19425284921181, 1.17788731755683, 
                -0.367897054652365 ), 
  `Nor4` = c(-0.840752912305256, 0.536548846040064, -0.277409459604357, 
                -0.241073614962264, -0.875313153342293, 1.61789645804321, 
                0.412287101096504, -1.11846661523232, -2.6274528854429, 
                -0.760452698231182), 
  `Tor1` = c(-0.968784779247286, -0.502809694119192, -0.231526399163731, 
                -0.530038395734114, -0.706006018337411, 3.58264357077653, 
                -0.127521010699219, 0.270523387217103, 1.68335644352003, 
                -0.314902131571829), 
  `Tor2` = c(-0.481754175843152, -0.440784040523259, -0.532975340622715, 
                -0.182089795101371, -0.564807490336052, 1.74119896504534, 
                -0.96169805631325, -0.721782763145306, -0.433459827401695, 
                -0.727495835245995 ), 
  `Tor3` = c(-0.889343429110847, 1.07937149728343, -0.215144871523998, 
                -0.92234350748557, -0.832108253417702, 2.02456082994848, 
                -0.0434322861759954, -0.523126561938426, -0.556984056084809, 
                -0.740331742513503), 
  `Tor4` = c(-0.858141567384178, 1.87728717064375, -0.381047638414538, 
                -0.613568289061259, -1.92838339196505, 2.23393705735665, 
                0.635389543483408, -0.466053620529111, -1.50483745357134, 
                -1.33400859143521), 
  `Tor5` = c(-0.486388736112514, 0.789390852922639, -0.869434195504952, 
                -0.70405854858187, -1.16488184095428, 2.91497178849082, 
                -2.10331904053714, -0.571130459068143, -0.219526004620518, 
                -0.301435496557957)
)

如何获取按列的 Wilcox.test 和 Fisher 提取文本，将 Nor1、Nor2、Nor3 和 Nor4 列与每行的 Tor1、Tor2、Tor3、Tor4 和 Tor5 列进行比较。然后，我想在最后一列添加两个测试的 p 值输出，得到 df2：

df2 <- data.frame( En_ID = c("KNT00000000003", "KNT00000000005", "KNT00000000419", "KNT00000000457", "KNT00000000460", "KNT00000000938", "KNT00000000971", "KNT00000001036", "KNT00000001084", "KNT00000001167" ), `Nor1` = c(-0.834165161710272, 1.02199443531549, -0.558658947885705, -0.390114219973209, -1.23551839713296, 3.11429434221998, 0.283932163407262, -1.16908518620064, -0.597054772455507, -0.593624543273255), `Nor2` = c(-1.18531035488942, 0.423719727339646, -1.23261719368372, 0.0855281133529292, -1.52366830232278, 3.36692586561211, 1.00323690950956, -0.000211248816114964, -4.74738483548391, -0.318176231083024), `Nor3` = c(-0.262659255267546, 1.3962481061442, -0.548673555705647, -0.0149651083306594, -1.45458689193089, 2.54126941463459, 1.17711308509307, -1.19425284921181, 1.17788731755683, -0.367897054652365 ), `Nor4` = c(-0.840752912305256, 0.536548846040064, -0.277409459604357, -0.241073614962264, -0.875313153342293, 1.61789645804321, 0.412287101096504, -1.11846661523232, -2.6274528854429, -0.760452698231182), `Tor1` = c(-0.968784779247286, -0.502809694119192, -0.231526399163731, -0.530038395734114, -0.706006018337411, 3.58264357077653, -0.127521010699219, 0.270523387217103, 1.68335644352003, -0.314902131571829), `Tor2` = c(-0.481754175843152, -0.440784040523259, -0.532975340622715, -0.182089795101371, -0.564807490336052, 1.74119896504534, -0.96169805631325, -0.721782763145306, -0.433459827401695, -0.727495835245995 ), `Tor3` = c(-0.889343429110847, 1.07937149728343, -0.215144871523998, -0.92234350748557, -0.832108253417702, 2.02456082994848, -0.0434322861759954, -0.523126561938426, -0.556984056084809, -0.740331742513503), `Tor4` = c(-0.858141567384178, 1.87728717064375, -0.381047638414538, -0.613568289061259, -1.92838339196505, 2.23393705735665, 0.635389543483408, -0.466053620529111, -1.50483745357134, -1.33400859143521), `Tor5` = c(-0.486388736112514, 0.789390852922639, -0.869434195504952, -0.70405854858187, -1.16488184095428, 2.91497178849082, -2.10331904053714, -0.571130459068143, -0.219526004620518, -0.301435496557957),`Tor4` = c(-0.858141567384178, 1.87728717064375, -0.381047638414538, -0.613568289061259, -1.92838339196505, 2.23393705735665, 0.635389543483408, -0.466053620529111, -1.50483745357134, -1.33400859143521), `p-value-wilcox` = c(0.8, 0.3, 0.7, 0.8, 0.9, 0.8, 0.7, -0.5, -0.7, -0.9), `p-value-fisher` = c(0.1, 0.7, 0.3, 0.1, 0.5, 0.3, 0.9, -0.2, -0.9, -0.4) )

在这里，我使用虚拟 p 值来提供所需输出的轮廓。真实数据有 >200 列，但两组（Nor 和 Tor）的样本数不相等。

我从堆栈中找到了一些示例，如下所述，并尝试复制它们，但惨遭失败。

威尔考克斯文本

费希尔提取文本

请帮我。

4

3 回答 3

1

fisher.test需要相同长度的数据，所以我假设你有相同数量的Nor和Tor列。

使用dplyr rowwise你可以使用 -

library(dplyr)

df1 %>%
  select(-Tor5) %>%
  rowwise() %>%
  mutate(p.value.wilcox = wilcox.test(c_across(starts_with('Nor')),
                              c_across(starts_with('Tor')))$p.value, 
         p.value.fisher = fisher.test(c_across(starts_with('Nor')),
                              c_across(starts_with('Tor')))$p.value)

或者在基础 R 中，使用apply-

nor_cols <- grep('Nor', names(df1))
tor_cols <- grep('Tor', names(df1))[-5]

cbind(df1, t(apply(df1[-1], 1, function(x) 
      c(p.value.wilcox = wilcox.test(x[nor_cols], x[tor_cols])$p.value,
        p.value.fisher = fisher.test(x[nor_cols], x[tor_cols])$p.value))))
于 2021-10-06T12:17:02.493 回答
1

Fisher 精确检验用于分类变量，我不确定您是否可以将其应用于连续数据。对于 Mann Whitney，您可以使用：

for (i in c(1:length(df1[,1]))){
  test_list <- wilcox.test(as.numeric(df1[i,c(2:5)]), as.numeric(df1[i,6:10]), exact = FALSE)
  df1[i,"p_val_MW"] <- test_list[[3]]
}
于 2021-10-06T12:25:09.983 回答
1

我们可以使用dapplyfrom collapsewhich 应该更快

library(collapse)
 cbind(df1, dapply(slt(df1, -c(1, 10)), MARGIN = 1,
    FUN = function(x) c(wilcox = wilcox.test(x[1:4], x[5:8])$p.value,
    fisher = fisher.test(x[1:4], x[5:8])$p.value)))
            En_ID       Nor1          Nor2        Nor3       Nor4       Tor1       Tor2        Tor3       Tor4       Tor5    wilcox fisher
1  KNT00000000003 -0.8341652 -1.1853103549 -0.26265926 -0.8407529 -0.9687848 -0.4817542 -0.88934343 -0.8581416 -0.4863887 0.6857143      1
2  KNT00000000005  1.0219944  0.4237197273  1.39624811  0.5365488 -0.5028097 -0.4407840  1.07937150  1.8772872  0.7893909 0.8857143      1
3  KNT00000000419 -0.5586589 -1.2326171937 -0.54867356 -0.2774095 -0.2315264 -0.5329753 -0.21514487 -0.3810476 -0.8694342 0.1142857      1
4  KNT00000000457 -0.3901142  0.0855281134 -0.01496511 -0.2410736 -0.5300384 -0.1820898 -0.92234351 -0.6135683 -0.7040585 0.1142857      1
5  KNT00000000460 -1.2355184 -1.5236683023 -1.45458689 -0.8753132 -0.7060060 -0.5648075 -0.83210825 -1.9283834 -1.1648818 0.3428571      1
6  KNT00000000938  3.1142943  3.3669258656  2.54126941  1.6178965  3.5826436  1.7411990  2.02456083  2.2339371  2.9149718 0.8857143      1
7  KNT00000000971  0.2839322  1.0032369095  1.17711309  0.4122871 -0.1275210 -0.9616981 -0.04343229  0.6353895 -2.1033190 0.1142857      1
8  KNT00000001036 -1.1690852 -0.0002112488 -1.19425285 -1.1184666  0.2705234 -0.7217828 -0.52312656 -0.4660536 -0.5711305 0.2000000      1
9  KNT00000001084 -0.5970548 -4.7473848355  1.17788732 -2.6274529  1.6833564 -0.4334598 -0.55698406 -1.5048375 -0.2195260 0.3428571      1
10 KNT00000001167 -0.5936245 -0.3181762311 -0.36789705 -0.7604527 -0.3149021 -0.7274958 -0.74033174 -1.3340086 -0.3014355 0.6857143      1
于 2021-10-06T17:20:28.767 回答