Question

我有一个长度为3的data.frame - sku_data：数据样本是：

sku_data <- data.frame(
  sku = c(1200337,1200518,1200586,1200590),
  units_current = I(list(c(1),1:3,1:4,2:4)),
  units_end = I(list(1:3,1,2:4,1:3))
)

#      sku units_current units_end
#1 1200337             1   1, 2, 3
#2 1200518       1, 2, 3         1
#3 1200586    1, 2, 3, 4   2, 3, 4
#4 1200590       2, 3, 4   1, 2, 3

我想删除units_current或units_end长度为<=1的元素。它们都是列表清单。

由于

Answer 1

怎么样：

f <- function(x) which(sapply(x, length) <= 1L)

sku_data[-unlist(lapply(sku_data[-1], f)), ]
#       sku units_current units_end
# 3 1200586    1, 2, 3, 4   2, 3, 4
# 4 1200590       2, 3, 4   1, 2, 3

在多个/不均匀长度上进行测试匹配

(sku2 <- rbind(sku_data, sku_data, sku_data[1,]))
#       sku units_current units_end
# 1 1200337             1   1, 2, 3
# 2 1200518       1, 2, 3         1
# 3 1200586    1, 2, 3, 4   2, 3, 4
# 4 1200590       2, 3, 4   1, 2, 3
# 5 1200337             1   1, 2, 3
# 6 1200518       1, 2, 3         1
# 7 1200586    1, 2, 3, 4   2, 3, 4
# 8 1200590       2, 3, 4   1, 2, 3
# 9 1200337             1   1, 2, 3
sku2[-unlist(lapply(sku2[-1], f)), ]
#       sku units_current units_end
# 3 1200586    1, 2, 3, 4   2, 3, 4
# 4 1200590       2, 3, 4   1, 2, 3
# 7 1200586    1, 2, 3, 4   2, 3, 4
# 8 1200590       2, 3, 4   1, 2, 3

Answer 2

另一种选择是：

sku_data[!rowSums(`dim<-`(rapply(sku_data[,-1], length) <=1,
                                         dim(sku_data)-c(0,1))),]
#      sku units_current units_end
#3 1200586    1, 2, 3, 4   2, 3, 4
#4 1200590       2, 3, 4   1, 2, 3

使用列表列表从data.frame中删除元素

2 个答案: