我如何创建一个for循环或一个列表推导式,以便每次迭代都给我两个元素?
l = [1,2,3,4,5,6]
for i,k in ???:
print str(i), '+', str(k), '=', str(i+k)
输出:
1+2=3
3+4=7
5+6=11
我如何创建一个for循环或一个列表推导式,以便每次迭代都给我两个元素?
l = [1,2,3,4,5,6]
for i,k in ???:
print str(i), '+', str(k), '=', str(i+k)
输出:
1+2=3
3+4=7
5+6=11
当前回答
可以使用more_itertools包。
import more_itertools
lst = range(1, 7)
for i, j in more_itertools.chunked(lst, 2):
print(f'{i} + {j} = {i+j}')
其他回答
虽然使用zip的所有答案都是正确的,但我发现自己实现功能会导致更可读的代码:
def pairwise(it):
it = iter(it)
while True:
try:
yield next(it), next(it)
except StopIteration:
# no more elements in the iterator
return
it = iter(it)部分确保它实际上是一个迭代器,而不仅仅是一个可迭代对象。如果它已经是一个迭代器,这一行是一个无操作。
用法:
for a, b in pairwise([0, 1, 2, 3, 4, 5]):
print(a + b)
另一种更清洁的解决方案
def grouped(itr, n=2):
itr = iter(itr)
end = object()
while True:
vals = tuple(next(itr, end) for _ in range(n))
if vals[-1] is end:
return
yield vals
更多定制选项
from collections.abc import Sized
def grouped(itr, n=2, /, truncate=True, fillvalue=None, strict=False, nofill=False):
if strict:
if isinstance(itr, Sized):
if len(itr) % n != 0:
raise ValueError(f"{len(itr)=} is not divisible by {n=}")
itr = iter(itr)
end = object()
while True:
vals = tuple(next(itr, end) for _ in range(n))
if vals[-1] is end:
if vals[0] is end:
return
if strict:
raise ValueError("found extra stuff in iterable")
if nofill:
yield tuple(v for v in vals if v is not end)
return
if truncate:
return
yield tuple(v if v is not end else fillvalue for v in vals)
return
yield vals
我希望这是一种更优雅的方法。
a = [1,2,3,4,5,6]
zip(a[::2], a[1::2])
[(1, 2), (3, 4), (5, 6)]
有很多方法可以做到这一点。例如:
lst = [1,2,3,4,5,6]
[(lst[i], lst[i+1]) for i,_ in enumerate(lst[:-1])]
>>>[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]
list(zip(*[iter(lst)]*2))
>>>[(1, 2), (3, 4), (5, 6)]
>>> l = [1,2,3,4,5,6]
>>> zip(l,l[1:])
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]
>>> zip(l,l[1:])[::2]
[(1, 2), (3, 4), (5, 6)]
>>> [a+b for a,b in zip(l,l[1:])[::2]]
[3, 7, 11]
>>> ["%d + %d = %d" % (a,b,a+b) for a,b in zip(l,l[1:])[::2]]
['1 + 2 = 3', '3 + 4 = 7', '5 + 6 = 11']