邮编还是用R枚举? [英] Zip or enumerate in R?

问题描述

这些Python列表推导的R等效项是什么:

What are the R equivalents for these Python list comprehensions:

[(i,j) for i,j in zip(index, Values)]
[(i,j) for i,j in enumerate(Values)]
[(i,j) for i,j in enumerate(range(10,20))]   %MWE, indexing or enumerating to 
                                            %keep up with the index, there may 
                                            %be some parameter to look this up

带有输出的示例

>>> [(i,j) for i,j in enumerate(range(10,20))]
[(0, 10), (1, 11), (2, 12), (3, 13), (4, 14), (5, 15), (6, 16), (7, 17), (8, 18), (9, 19)]

我已经用R中的一些技巧解决了这个问题，但是已经不记得了，第一个想法是itertools -pkg，但是我希望找到一种更惯用的做事方式.

I have solved this problem earlier with some trick in R but cannot remember anymore, the first idea was itertools -pkg but I am hoping to find a more idiomatic way of doing things.

推荐答案

Python enumerate的答案:

Answer for python enumerate:

在R中，列表是有序的(请参见此答案 ).因此，您只需要索引键(使用names()[i])或值(使用[[i]]).

In R, a list is ordered (see this answer). Thus, all you need is to index either keys (using names()[i]) or values (using [[i]]).

使用seq_along(或者可以做for(i in 1:length(mylist)){...}):

> mylist <- list('a'=10,'b'=20,'c'=30)
> for (i in seq_along(mylist)){
+   print(paste(i,names(mylist)[i],mylist[[i]]))
+ }
[1] "1 a 10"
[1] "2 b 20"
[1] "3 c 30"

python zip的答案:

Answer for python zip:

请参阅上述答案之一，以模仿元组列表.我更喜欢BondedDust的答案所示的数据框:

See one of the above answers to mimic the list of tuples. My preference is towards a data frame as shown in BondedDust's answer:

> x <- 1:3
> y <- 4:6
> data.frame(x=x, y=y)
  x y
1 1 4
2 2 5
3 3 6

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