Question

我有谁列出：

list_a = [['abc', 10],['def', 20],['ghj', 30]]
list_b = [['abc', 40],['def', 50],['ghj', 60],['klm', 70]]

我想通过list.a浏览list.b所以我可以删除备用元素＆＃34; list.b [3]＆＃34;并使用列表成员计算：

list_a[0][1] - list_b[0][1] = -30
list_a[1][1] - list_b[1][1] = -30
list_a[2][1] - list_b[2][1] = -30

我到底该怎么做？

Answer 1

使用#example data of what I want dat$group <- c(1, 1, 1, 2, 2, 1, 1, 1, 2, 2) #this is the variable I want dat$seq <- with(dat, ave(as.character(category), category, FUN = seq_along)) #count sequence variable, which is different than what I'm after because it does not repeat within sequential categories > dat id time category group seq 1 a 2016-10-24 21:05:00 low 1 1 2 a 2016-10-24 21:10:00 high 1 1 3 a 2016-10-24 21:15:00 high 1 2 4 a 2016-10-24 21:20:00 low 2 2 5 a 2016-10-24 21:25:00 low 2 3 6 b 2016-10-24 21:30:00 low 1 4 7 b 2016-10-24 21:35:00 high 1 3 8 b 2016-10-24 21:40:00 high 1 4 9 b 2016-10-24 21:45:00 low 2 5 10 b 2016-10-24 21:50:00 low 2 6。它将自动修剪到最短迭代的长度。这是列表理解：

zip

浏览多维列表

1 个答案: