如何对 R 中的组按列执行 wilcox.test
How to perform column-wise wilcox.test on groups in R
我有 data.frame df1:
set.seed(12345)
df1 <- data.frame(group=c(rep("apple", 4), rep("pear",6)), a=rnorm(10,0,0.4),
b=rnorm(10,0,0.2),
c=rnorm(10,0,0.7), d=rnorm(10,0,0.9), e=rnorm(10,0,0.5))
如何获得比较苹果（行 1:4）和梨（行 5:10）的列 wilcox.test p 值并将该 p 值添加到新行底部，导致 df2:
df2 <- data.frame(group=c(rep("apple", 4), rep("pear",6), "wilcox.test"),
a=c(rnorm(10,0,0.4), 0.393768635), b=c(rnorm(10,0,0.2), 0.286422023),
c=c(rnorm(10,0,0.7), 1), d=c(rnorm(10,0,0.9), 0.033006258),
e=c(rnorm(10,0,0.5), 1))
> df2
group a b c d e
1 apple 0.23421153 -0.02324956 0.5457353 0.73068586 0.5642554
2 apple 0.28378641 0.36346241 1.0190496 1.97715019 -1.190179
3 apple -0.04372133 0.07412557 -0.4510299 1.8442713 -0.5301328
4 apple -0.18139887 0.10404329 -1.0871962 1.46920108 0.4685703
5 pear 0.24235498 -0.1501064 -1.1183967 0.22884407 0.4272259
6 pear -0.72718239 0.16337997 1.2635683 0.44206945 0.7303647
7 pear 0.25203942 -0.1772715 -0.3371532 -0.29167792 -0.7065494
8 pear -0.11047364 -0.06631552 0.4342659 -1.49584522 0.2837016
9 pear -0.1136639 0.22414253 0.4284864 1.59096047 0.2915938
10 pear -0.3677288 0.05974474 -0.1136177 0.02322094 -0.6533994
11 wilcox.test 0.393768635 0.286422023 1 0.033006258 1
这是一个基本的 R 选项 -
subset 其中两个值的数据，使用 Map 对每一列应用 wilcox.test 并从中提取 p 值，将其作为新行添加到已经存在的 df1.
rbind(df1, data.frame(group = 'wilcox.test',
mapply(function(x, y) wilcox.test(x, y)$p.value,
subset(df1, group == 'apple', select = -group),
subset(df1, group == 'pear', select = -group)) |>
t() |> data.frame()))
# group a b c d e
#1 apple 0.23421153 -0.02324956 0.5457353 0.73068586 0.5642554
#2 apple 0.28378641 0.36346241 1.0190496 1.97715019 -1.1901790
#3 apple -0.04372133 0.07412557 -0.4510299 1.84427130 -0.5301328
#4 apple -0.18139887 0.10404329 -1.0871962 1.46920108 0.4685703
#5 pear 0.24235498 -0.15010640 -1.1183967 0.22884407 0.4272259
#6 pear -0.72718239 0.16337997 1.2635683 0.44206945 0.7303647
#7 pear 0.25203942 -0.17727150 -0.3371532 -0.29167792 -0.7065494
#8 pear -0.11047364 -0.06631552 0.4342659 -1.49584522 0.2837016
#9 pear -0.11366390 0.22414253 0.4284864 1.59096047 0.2915938
#10 pear -0.36772880 0.05974474 -0.1136177 0.02322094 -0.6533994
#11 wilcox.test 0.47619048 0.35238095 1.0000000 0.03809524 1.0000000
使用 R 4.1 中的管道 (|>) 以提高可读性。
我们可以用 dplyr 来做。我们可以用summarise和wilcox.test，提取p.value和$，然后用bind_rows绑定p.values并调整p.values作为最后一行。
df1 %>% summarise(across(!group, ~wilcox.test(.x ~ group)$p.value)) %>%
bind_rows(., p.adjust(., method = 'bonferroni')) %>%
bind_rows(df1, .) %>%
mutate(group=replace(group, is.na(group), c('p.values', 'adjusted_p.values')))
group a b c d e
1 apple 0.23421153 -0.02324956 0.5457353 0.73068586 0.5642554
2 apple 0.28378641 0.36346241 1.0190496 1.97715019 -1.1901790
3 apple -0.04372133 0.07412557 -0.4510299 1.84427130 -0.5301328
4 apple -0.18139887 0.10404329 -1.0871962 1.46920108 0.4685703
5 pear 0.24235498 -0.15010640 -1.1183967 0.22884407 0.4272259
6 pear -0.72718239 0.16337997 1.2635683 0.44206945 0.7303647
7 pear 0.25203942 -0.17727150 -0.3371532 -0.29167792 -0.7065494
8 pear -0.11047364 -0.06631552 0.4342659 -1.49584522 0.2837016
9 pear -0.11366390 0.22414253 0.4284864 1.59096047 0.2915938
10 pear -0.36772880 0.05974474 -0.1136177 0.02322094 -0.6533994
11 p.values 0.47619048 0.35238095 1.0000000 0.03809524 1.0000000
12 adjusted_p.values 1.00000000 1.00000000 1.0000000 0.19047619 1.0000000
我有 data.frame df1:
set.seed(12345)
df1 <- data.frame(group=c(rep("apple", 4), rep("pear",6)), a=rnorm(10,0,0.4),
b=rnorm(10,0,0.2),
c=rnorm(10,0,0.7), d=rnorm(10,0,0.9), e=rnorm(10,0,0.5))
如何获得比较苹果（行 1:4）和梨（行 5:10）的列 wilcox.test p 值并将该 p 值添加到新行底部，导致 df2:
df2 <- data.frame(group=c(rep("apple", 4), rep("pear",6), "wilcox.test"),
a=c(rnorm(10,0,0.4), 0.393768635), b=c(rnorm(10,0,0.2), 0.286422023),
c=c(rnorm(10,0,0.7), 1), d=c(rnorm(10,0,0.9), 0.033006258),
e=c(rnorm(10,0,0.5), 1))
> df2
group a b c d e
1 apple 0.23421153 -0.02324956 0.5457353 0.73068586 0.5642554
2 apple 0.28378641 0.36346241 1.0190496 1.97715019 -1.190179
3 apple -0.04372133 0.07412557 -0.4510299 1.8442713 -0.5301328
4 apple -0.18139887 0.10404329 -1.0871962 1.46920108 0.4685703
5 pear 0.24235498 -0.1501064 -1.1183967 0.22884407 0.4272259
6 pear -0.72718239 0.16337997 1.2635683 0.44206945 0.7303647
7 pear 0.25203942 -0.1772715 -0.3371532 -0.29167792 -0.7065494
8 pear -0.11047364 -0.06631552 0.4342659 -1.49584522 0.2837016
9 pear -0.1136639 0.22414253 0.4284864 1.59096047 0.2915938
10 pear -0.3677288 0.05974474 -0.1136177 0.02322094 -0.6533994
11 wilcox.test 0.393768635 0.286422023 1 0.033006258 1
这是一个基本的 R 选项 -
subset 其中两个值的数据，使用 Map 对每一列应用 wilcox.test 并从中提取 p 值，将其作为新行添加到已经存在的 df1.
rbind(df1, data.frame(group = 'wilcox.test',
mapply(function(x, y) wilcox.test(x, y)$p.value,
subset(df1, group == 'apple', select = -group),
subset(df1, group == 'pear', select = -group)) |>
t() |> data.frame()))
# group a b c d e
#1 apple 0.23421153 -0.02324956 0.5457353 0.73068586 0.5642554
#2 apple 0.28378641 0.36346241 1.0190496 1.97715019 -1.1901790
#3 apple -0.04372133 0.07412557 -0.4510299 1.84427130 -0.5301328
#4 apple -0.18139887 0.10404329 -1.0871962 1.46920108 0.4685703
#5 pear 0.24235498 -0.15010640 -1.1183967 0.22884407 0.4272259
#6 pear -0.72718239 0.16337997 1.2635683 0.44206945 0.7303647
#7 pear 0.25203942 -0.17727150 -0.3371532 -0.29167792 -0.7065494
#8 pear -0.11047364 -0.06631552 0.4342659 -1.49584522 0.2837016
#9 pear -0.11366390 0.22414253 0.4284864 1.59096047 0.2915938
#10 pear -0.36772880 0.05974474 -0.1136177 0.02322094 -0.6533994
#11 wilcox.test 0.47619048 0.35238095 1.0000000 0.03809524 1.0000000
使用 R 4.1 中的管道 (|>) 以提高可读性。
我们可以用 dplyr 来做。我们可以用summarise和wilcox.test，提取p.value和$，然后用bind_rows绑定p.values并调整p.values作为最后一行。
df1 %>% summarise(across(!group, ~wilcox.test(.x ~ group)$p.value)) %>%
bind_rows(., p.adjust(., method = 'bonferroni')) %>%
bind_rows(df1, .) %>%
mutate(group=replace(group, is.na(group), c('p.values', 'adjusted_p.values')))
group a b c d e
1 apple 0.23421153 -0.02324956 0.5457353 0.73068586 0.5642554
2 apple 0.28378641 0.36346241 1.0190496 1.97715019 -1.1901790
3 apple -0.04372133 0.07412557 -0.4510299 1.84427130 -0.5301328
4 apple -0.18139887 0.10404329 -1.0871962 1.46920108 0.4685703
5 pear 0.24235498 -0.15010640 -1.1183967 0.22884407 0.4272259
6 pear -0.72718239 0.16337997 1.2635683 0.44206945 0.7303647
7 pear 0.25203942 -0.17727150 -0.3371532 -0.29167792 -0.7065494
8 pear -0.11047364 -0.06631552 0.4342659 -1.49584522 0.2837016
9 pear -0.11366390 0.22414253 0.4284864 1.59096047 0.2915938
10 pear -0.36772880 0.05974474 -0.1136177 0.02322094 -0.6533994
11 p.values 0.47619048 0.35238095 1.0000000 0.03809524 1.0000000
12 adjusted_p.values 1.00000000 1.00000000 1.0000000 0.19047619 1.0000000