r - 如何使用 dplyr 计算与 rowmean 的比率

Question

我有一个数据框，其中包含不同样本中基因的定量值，我想将每个值除以行平均值。后跟所有值的 log2。这可以通过 base R 完成，如下所示，但我无法使用管道使其工作。

示例数据框:

df <- data.frame("Gene_Symbol" = c("Gene1","Gene2","Gene3","Gene4","Gene5","Gene6","Gene7"), 
             "Sample1" = c(85657.97656,54417.78906,110949.3281,53197.45313,87156.80469,NA,23880.2832), 
             "Sample2" = c(10423.40918,41660.73047,40094.54688,49519.78125,129387.1094,NA,23903.25977), 
             "Sample3" = c(18778.68359,43655.79688,NA,57447.08984,113266.1484,44810.26172,26316.6543), 
             "Sample4" = c(23919.53125,47829.02344,NA,51478.58203,116275.3359,43110.94922,25417.45508), 
             "Sample5" = c(NA,46677.20313,63389.45313,48722.15234,NA,77135.52344,40265.6875), 
             "Sample6" = c(NA,68596.22656,56802.60938,44712.64063,NA,47744.17969,33689.62891), 
             "Sample7" = c(NA,80506.14844,48722.99219,38629.00781,NA,37885,36638.02344))

想要得到 log2 的比率与 rowmean，如下所示:

  Gene_Symbol      Sample1     Sample2     Sample3     Sample4     Sample5     Sample6    Sample7
1       Gene1  1.303863983 -1.73489640 -0.88562768 -0.53653450          NA          NA         NA
2       Gene2 -0.009130358 -0.39452056 -0.32703546 -0.19532236 -0.23049058  0.32492052  0.5558903
3       Gene3  0.793942295 -0.67448070          NA          NA -0.01364391 -0.17192953 -0.3932840
4       Gene4  0.115606000  0.01225376  0.22648263  0.06822114 -0.01117331 -0.13506843 -0.3460666
5       Gene5 -0.355634714  0.21437397  0.02239683  0.06022518          NA          NA         NA
6       Gene6           NA          NA -0.16205178 -0.21782661  0.62151449 -0.07055606 -0.4042542
7       Gene7 -0.329904867 -0.32851744 -0.18974873 -0.23990523  0.42382615  0.16657972  0.2876169

用基数 R 计算 rowMeans

rowMeanValues <- rowMeans(df[,2:ncol(df)], na.rm = TRUE)

将所有量化值除以 rowMeanValues

df[,2:ncol(df)] <- sweep(df[,2:ncol(df)],
                                     MARGIN = 1, FUN = "/",
                                     STATS = rowMeanValues)

比率的 log2

df[,2:ncol(df)] <- log2(df[,2:ncol(df)])

这给了我上面想要的表格。我如何在 dplyr 中进行这些计算？

在下面尝试过，但它是按列平均值而不是行平均值划分的

df %>% mutate_at(vars(starts_with('Sample')), funs(./mean(., na.rm = TRUE)))

感谢帮助!亨里克

Answer 1

一个选项是先计算 rowMeans 并将其创建为列，然后在下一步中执行 mutate_at。在这里，我们使用 base R 中的 rowMeans，因为它比 rowwise 或其他形式或 reshape 计算行方式均值更有效

library(dplyr)
df %>%
   mutate(Mean = rowMeans(select(., starts_with('Sample')), na.rm = TRUE)) %>% 
   mutate_at(vars(starts_with('Sample')), ~ log2(./Mean)) %>%
   select(-Mean) # removing the Mean column from the dataset
#Gene_Symbol      Sample1     Sample2     Sample3     Sample4     Sample5     Sample6    Sample7
#1       Gene1  1.303863983 -1.73489640 -0.88562768 -0.53653450          NA          NA         NA
#2       Gene2 -0.009130358 -0.39452056 -0.32703546 -0.19532236 -0.23049058  0.32492052  0.5558903
#3       Gene3  0.793942295 -0.67448070          NA          NA -0.01364391 -0.17192953 -0.3932840
#4       Gene4  0.115606000  0.01225376  0.22648263  0.06822114 -0.01117331 -0.13506843 -0.3460666
#5       Gene5 -0.355634714  0.21437397  0.02239683  0.06022518          NA          NA         NA
#6       Gene6           NA          NA -0.16205178 -0.21782661  0.62151449 -0.07055606 -0.4042542
#7       Gene7 -0.329904867 -0.32851744 -0.18974873 -0.23990523  0.42382615  0.16657972  0.2876169

此外，mutate_at 中的 . 是实际的列值，因此采用 的 mean。只会按列进行平均而不按行进行