Question

我有一个数据框：

df <- structure(list(gene = structure(1:6, .Label = c("128up", "14-3-3epsilon", 
"14-3-3zeta", "140up", "18SrRNA-Psi:CR41602", "18SrRNA-Psi:CR45861"
), class = "factor"), fpkm = list(NULL, 0.4, NA_real_, NULL, 
    NULL, NULL)), .Names = c("gene", "fpkm"), row.names = c(NA, 
6L), class = "data.frame")

                 gene fpkm
1               128up NULL
2       14-3-3epsilon  0.4
3          14-3-3zeta   NA
4               140up NULL
5 18SrRNA-Psi:CR41602 NULL
6 18SrRNA-Psi:CR45861 NULL

我想根据level中的值添加新列fpkm。如果值为NULL或NA，我希望值为'not_expressed , else表达式。

我正在使用mutate来实现此目的，分别为NA和NULL值，但这对NULL值没有预期效果：

mutate(df, level = ifelse(is.na(fpkm), 'not_expressed' , 'expressed'))

                 gene fpkm         level
1               128up NULL     expressed
2       14-3-3epsilon  0.4     expressed
3          14-3-3zeta   NA not_expressed # Expected
4               140up NULL     expressed
5 18SrRNA-Psi:CR41602 NULL     expressed
6 18SrRNA-Psi:CR45861 NULL     expressed

  mutate(df, level = ifelse(is.null(fpkm), 'not_expressed' , 'expressed'))

                 gene fpkm     level
1               128up NULL expressed
2       14-3-3epsilon  0.4 expressed
3          14-3-3zeta   NA expressed
4               140up NULL expressed
5 18SrRNA-Psi:CR41602 NULL expressed
6 18SrRNA-Psi:CR45861 NULL expressed

我无法弄清楚为什么这不起作用 - is.null(unlist(test$fpkm[1]))会返回TRUE

我也尝试过： ifelse(is.null(df$fpkm), 'not_expressed', 'expressed')

和： ifelse(is.null(unlist(df$fpkm)), 'not_expressed', 'expressed')

......两者都不起作用

Answer 1

Base R对此非常好：

> df$level <-  ifelse( df$fpkm == 'NULL' | is.na(df$fpkm), 'not_expressed', 'expressed')
> df
                 gene fpkm         level
1               128up NULL not_expressed
2       14-3-3epsilon  0.4     expressed
3          14-3-3zeta   NA not_expressed
4               140up NULL not_expressed
5 18SrRNA-Psi:CR41602 NULL not_expressed
6 18SrRNA-Psi:CR45861 NULL not_expressed

Answer 2

最后一句话是您问题的线索。您必须unlist才能获得正确的结果

unlist(test$fpkm[1])

fpkm被保存为数据框中的列表（而不是矢量，这是典型的）

'data.frame':   6 obs. of  2 variables:
 $ gene: Factor w/ 6 levels "128up","14-3-3epsilon",..: 1 2 3 4 5 6
 $ fpkm:List of 6
  ..$ : NULL
  ..$ : num 0.4
  ..$ : num NA
  ..$ : NULL
  ..$ : NULL
  ..$ : NULL

您可以使用

获得正确的结果

  mutate(df, level = ifelse(map_lgl(df$fpkm, ~is.null(.x) || is.na(.x)), 'not_expressed' , 'expressed'))

Answer 3

将 NULL 转换为 NA ，然后使用ifelse：

# Convert NULL to NA
df$fpkm[ sapply(df$fpkm, is.null) ] <- NA

# I would also drop the list, it is up to you.
# df$fpkm <- unlist(df$fpkm)

# Then use ifelse as usual
df$level <-  ifelse(is.na(df$fpkm), "not_expressed", "expressed")

# result
df
#                  gene fpkm         level
# 1               128up   NA not_expressed
# 2       14-3-3epsilon  0.4     expressed
# 3          14-3-3zeta   NA not_expressed
# 4               140up   NA not_expressed
# 5 18SrRNA-Psi:CR41602   NA not_expressed
# 6 18SrRNA-Psi:CR45861   NA not_expressed

Answer 4

作为替代解决方案，您可以尝试找出列值的长度。代替：

mutate(df, level = ifelse(is.na(fpkm), 'not_expressed' , 'expressed'))

尝试一下：

library(stringr)
mutate(df, level = if_else(str_length(fpkm)>0, 'expressed','not_expressed'))

使用ifelse和is.null来测试NULL值

4 个答案: