英文:
how to iterate through a list of tuples with lists and group elements sequentially?
问题
我有这个列表lis = [(0, [75, 1, 30]), (1, [41, 49, 55]), (2, [28, 53, 45])]我试图将这些值按顺序分组在元组中。有什么方法可以做到这一点？期望的输出是[(0,75), (75, 1), (1, 30), (1, 41), (41,49), (49,55), (2,28), (28, 53), (53, 45)]found = []for tup in edge_list:found.append((tup[0], tup[1][0]))found.append((tup[1][0], tup[1][1]))found.append((tup[1][1], tup[1][2]))print(found)是否有更好/更简单的方法来迭代这个列表，不管第二个元组的大小是多少[(0, [82, 70, 79, 77, 42]), (1, [40, 61, 58, 66, 69]), (2, [80, 30, 12, 77, 9])]
英文:
I have this list
lis = [(0, [75, 1, 30]), (1, [41, 49, 55]), (2, [28, 53, 45])]
and I'm trying to group these values sequentially together in tuples. What would be a way to do this?
Expected output is
[(0,75), (75, 1), (1, 30), (1, 41), (41,49), (49,55), (2,28), (28, 53), (53, 45)]found = []for tup in edge_list:found.append((tup[0], tup[1][0]))found.append((tup[1][0], tup[1][1]))found.append((tup[1][1], tup[1][2]))print(found)
is there a better/easier approach to iterate over this no matter what the size of the second tuple looks like
[(0, [82, 70, 79, 77, 42]), (1, [40, 61, 58, 66, 69]), (2, [80, 30, 12, 77, 9])]
答案1
得分: 1
通过*运算符展开lis中的元素是一个有用的技巧：
lis = [(0, [75, 1, 30]), (1, [41, 49, 55]), (2, [28, 53, 45])][(a, *b) for a, b in lis][(0, 75, 1, 30), (1, 41, 49, 55), (2, 28, 53, 45)]
给定任何序列，你可以通过将其与自身的偏移版本进行zip处理来产生你想要的配对：
list(zip(s, s[1:]))[(1, 2), (2, 3), (3, 4)]
结合这两种技巧（即取出lis中的序列，然后将它们转换为配对），可以进行如下操作：
[list(zip((a, *b), b)) for a, b in lis][[(0, 75), (75, 1), (1, 30)], [(1, 41), (41, 49), (49, 55)], [(2, 28), (28, 53), (53, 45)]]
为了将它们全部放入一个配对的列表中，可以使用嵌套的列表推导式：
[z for a, b in lis for z in zip((a, *b), b)][(0, 75), (75, 1), (1, 30), (1, 41), (41, 49), (49, 55), (2, 28), (28, 53), (53, 45)]
英文:
A useful trick is to flatten out the elements of lis via the * operator:
>>> lis = [(0, [75, 1, 30]), (1, [41, 49, 55]), (2, [28, 53, 45])]>>> [(a, *b) for a, b in lis][(0, 75, 1, 30), (1, 41, 49, 55), (2, 28, 53, 45)]
Given any sequence, you can produce the kind of pairing you're looking for by zipping it with an offset version of itself:
>>> list(zip(s, s[1:]))[(1, 2), (2, 3), (3, 4)]
Combining those two techniques (i.e. taking the sequences within lis and then converting them into pairs), you can do:
>>> [list(zip((a, *b), b)) for a, b in lis][[(0, 75), (75, 1), (1, 30)], [(1, 41), (41, 49), (49, 55)], [(2, 28), (28, 53), (53, 45)]]
and to get them all into a single list of pairs, we can do a nested list comprehension:
>>> [z for a, b in lis for z in zip((a, *b), b)][(0, 75), (75, 1), (1, 30), (1, 41), (41, 49), (49, 55), (2, 28), (28, 53), (53, 45)]
答案2
得分: 1
使用列表解包和 zip 方法lis = [(0, [75, 1, 30]), (1, [41, 49, 55]), (2, [28, 53, 45])]result = []for a, b in lis:res = [a, *b]for i, j in zip(res, res[1:]):result.append((i,j))result# 输出如下：[(0, 75),(75, 1),(1, 30),(1, 41),(41, 49),(49, 55),(2, 28),(28, 53),(53, 45)]
英文:
using list unpacking and zip method
In [49]: lis = [(0, [75, 1, 30]), (1, [41, 49, 55]), (2, [28, 53, 45])]In [50]: result = []...: for a, b in lis:...: res = [a, *b]...: for i, j in zip(res, res[1:]):...: result.append((i,j))...:In [51]: resultOut[51]:[(0, 75),(75, 1),(1, 30),(1, 41),(41, 49),(49, 55),(2, 28),(28, 53),(53, 45)]
答案3
得分: 0
以下是代码部分的中文翻译：
import itertoolslis = [(0, [75, 1, 30]), (1, [41, 49, 55]), (2, [28, 53, 45])]found = [*itertools.chain(*(itertools.pairwise(a) for *a, a[1:] in lis))]print(found)
英文:
Having fun with itertools...
from itertools import pairwise, chainlis = [(0, [75, 1, 30]), (1, [41, 49, 55]), (2, [28, 53, 45])]found = [*chain(*(pairwise(a) for *a, a[1:] in lis))]print(found)
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