我对Python很陌生,我仍然很难将语言本身用于我的程序。这就是我到目前为止所拥有的:
# Purpose: 'twolists' = takes 2 lists, & returns a new list containing
# alternating elements of lists.
# Return = final_list
# Parameter = list1, list2
def twolists(list1, list2): # don't forget to return final_list
alt_list = []
a1 = len(list1)
a2 = len(list2)
for i in range(# ? ):
# append one thing from list1 to alt_list - How?
# append one thing from list2 to alt_list - How?
现在该程序应该产生如下输出:
outcome = twolists([ ], ['w', 'x', 'y', 'z'])
print(outcome)
['w', 'x', 'y', 'z']
outcome = twolists([0, 1], ['w', 'x'])
print(outcome)
[0, 'w', 1, 'x']
outcome = twolists([0, 1], ['w', 'x', 'y', 'z'])
print(outcome)
[0, 'w', 1, 'x', 'y', 'z']
outcome = twolists([0, 1, 2, 3], ['w', 'x'])
print(outcome)
[0, 'w', 1, 'x', 2, 3]
答案 0 :(得分:6)
def twolists(list1, list2):
newlist = []
a1 = len(list1)
a2 = len(list2)
for i in range(max(a1, a2)):
if i < a1:
newlist.append(list1[i])
if i < a2:
newlist.append(list2[i])
return newlist
答案 1 :(得分:5)
这使用来自itertools的zip_longest
(它是标准库的一部分)来组成列表理解,以将两个列表中的项目交错为tuple
,默认情况下使用{{1}填充值。
这也使用来自None
的{{1}}来展平列表。
最后,它会过滤列表中的chain
项:
itertools
或者按照@EliKorvigo的建议,使用None
进行懒惰迭代:
from itertools import chain, zip_longest
def twolists(l1, l2):
return [x for x in chain(*zip_longest(l1, l2)) if x is not None]
测试
itertools.chain.from_iterable
答案 2 :(得分:3)
基本方法:
您可以正常zip()
列表,如果两个列表的大小不同,则附加最大列表的其余部分:
def two_lists(lst1, lst2):
result = []
for pair in zip(lst1, lst2):
result.extend(pair)
if len(lst1) != len(lst2):
lsts = [lst1, lst2]
smallest = min(lsts, key = len)
biggest = max(lsts, key = len)
rest = biggest[len(smallest):]
result.extend(rest)
return result
其工作原理如下:
>>> print(two_lists([], ['w', 'x', 'y', 'z']))
['w', 'x', 'y', 'z']
>>> print(two_lists([0, 1], ['w', 'x']))
[0, 'w', 1, 'x']
>>> print(two_lists([0, 1], ['w', 'x', 'y', 'z']))
[0, 'w', 1, 'x', 'y', 'z']
>>> print(two_lists([0, 1, 2, 3], ['w', 'x']))
[0, 'w', 1, 'x', 2, 3]
另一种可能的方法:
您也可以使用collections.deque
预先将列表转换为deque()
个对象,并使用popleft()
弹出每个对象的开头,直到其中一个对象为空。然后你可以追加尚未为空的列表的其余部分。
以下是一个例子:
def two_lists2(lst1, lst2):
result = []
fst, snd = deque(lst1), deque(lst2)
while fst and snd:
result.append(fst.popleft())
result.append(snd.popleft())
rest = leftover(fst, snd)
if rest:
result.extend(rest)
return result
def leftover(x, y):
if x and not y:
return x
elif y and not x:
return y
return None
注意:这两种方法都是O(n)
时间,这是这类问题的预期。
答案 3 :(得分:1)
def CombineLists(lst1, lst2):
return [item for x in zip(lst1,lst2) for item in x] + /
(lst2[len(lst1):] if len(lst2)>len(lst1) else lst1[len(lst2):])
答案 4 :(得分:-1)
这是一个处理迭代器的解决方案。这样做的好处是它可以与任何可迭代的数据结构一起使用,而不仅仅是列表。
def twolists(list1, list2):
result = []
iter1 = iter(list1)
iter2 = iter(list2)
try:
while True:
result.append(next(iter1))
result.append(next(iter2))
except StopIteration:
# This exception will be raised when either of the iterators
# hits the end of the sequence.
pass
# One of the lists is exhausted, but not both of them. We need
# to finish exhausting the lists.
try:
while True:
result.append(next(iter1))
except StopIteration:
pass
try:
while True:
result.append(next(iter2))
except StopIteration:
pass
return result
答案 5 :(得分:-1)
如果您不关心原始列表(以下示例中的a
和b
)是否发生更改,则可以使用以下代码段:
def twolists(a, b):
result = []
while len(a) > 0:
result.append(a.pop(0))
if len(b) > 0:
result.append(b.pop(0))
result += a + b
return result
twolists([ ], ['w', 'x', 'y', 'z'])
print(outcome)
outcome = twolists([0, 1], ['w', 'x'])
print(outcome)
outcome = twolists([0, 1], ['w', 'x', 'y', 'z'])
print(outcome)
outcome = twolists([0, 1, 2, 3], ['w', 'x'])
print(outcome)
产生以下输出:
['w', 'x', 'y', 'z']
[0, 'w', 1, 'x']
[0, 'w', 1, 'x', 'y', 'z']
[0, 'w', 1, 'x', 2, 3]