繁体   English   中英

特定列的行相关

[英]rowwise correlation for specific columns

对于所有行(符号),我想将五列F1_G-F5_G与五列F1_P-F5_P相关联。 这应该为每个符号给出一个相关值。

       Symbol       F1_G      F2_G       F3_G       F4_G       F5_G     F1_P     F2_P     F3_P     F4_P     F5_P
1       abca2  0.7696639  1.301428  0.8447565  0.6936672  0.6987410 9.873610 9.705205 8.044027 8.311364 9.961380
2        aco2  7.4274715  7.234892  7.8543164  8.0142983  8.1512194 9.620114 9.767721 7.607115 7.854960 9.472660
3       adat3 -2.0560126 -1.536868 -0.4181457 -1.1946602 -0.7707472 8.975871 8.645235 7.926262 7.432755 8.633583
4       adat3 -2.0560126 -1.536868 -0.4181457 -1.1946602 -0.7707472 9.620114 9.237699 7.162386 7.972086 8.872763
5        adnp  1.4228436  0.932214  0.8964153  0.8125162  0.9921002 9.177645 9.323443 8.507508 8.080413 8.633583
6     arhgap8 -2.6517712 -2.067164 -1.4918958 -2.6517712 -1.5474257 9.395681 8.861322 8.333381 8.038053 8.872763

我试过这样的事情,但并不考虑每一行:

res <- outer(df[, c(2,3,4,5,6)], df[, c(7,8,9,10,11)], function(X, Y){
  mapply(function(...) cor.test(..., na.action = "na.exclude")$estimate,
         X, Y)
})

出:

 Symbol  Cor
 abca2  0.14
 aco2   0.12

由于您想按行进行操作,因此我们可以在MARGIN = 1使用apply

#Get column indices ending with G
g_cols <- grep("G$", names(df))
#Get column indices ending with P
p_cols <- grep("P$", names(df))

apply(df, 1, function(x) cor.test(as.numeric(x[g_cols]), 
         as.numeric(x[p_cols]), na.action = "na.exclude")$estimate)

#       1        2        3        4        5        6 
# 0.21890 -0.52925 -0.52776 -0.82073  0.60473 -0.11785 

tidyverse方法将是

library(tidyverse)

df %>%
  mutate(row = row_number()) %>%
  select(-Symbol) %>%
  gather(key, value, -row) %>%
  group_by(row) %>%
  summarise(ans = cor.test(value[key %in% g_cols], value[key %in% p_cols], 
                  na.action = "na.exclude")$estimate)


#    row    ans
#   <int>  <dbl>
#1     1  0.219
#2     2 -0.529
#3     3 -0.528
#4     4 -0.821
#5     5  0.605
#6     6 -0.118

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM