Question

我有两个固定大小的字符串列表，我想检查这两个列表是否与以下示例“相似”：

list1 = ["a", None, "c", None, "e", None]
list2 = ["a", "b", "c", "d", "e", "f"]
similar = True
for i in xrange(6):
    if list1[i] is not None:
        if list1[i] != list2[i]:
            similar = False
            break

有没有更快的方法呢？

更新：我刚刚使用zip测试了一些解决方案。它们并不快，因为zip会考虑两个列表中的所有元素。请注意，在许多情况下，两个列表中的第一个元素是不同的，因此我提供的程序会立即停止检查剩余元素，从而提供更快的解决方案。

Answer 1

如果你担心提前退出，你可以这样做：

iter1, iter2 = iter(list1), iter(list2)
similar = True
while True:  # or `while similar`
    try:
        a, b = next(iter1), next(iter2)
    except StopIteration:
        break
    if a is not None and b is not None and a != b:
        similar = False
        break
# similar is your result

与...相同：

import itertools

all(a==b for a,b in itertools.izip(list1, list2) if a is not None and b is not None)

与...相同：

import itertools

for a,b in itertools.izip(list1, list2):
    if a is None or b is None:
        continue
    if a != b:
        break
else:
    # similar

Answer 2

有没有更快的方法呢？

是的，摆脱嵌套的if会产生一些小的差异：

list1 = ["a", None, "c", None, "e", None]
list2 = ["a", "b", "c", "d", "e", "f"]
similar = True
for i in xrange(6):
    if list1[i] is not None and list1[i] != list2[i]:
        similar = False
        break

print(similar)
# True

我认为这个算法错了。如果list1和list2互换：

list1, list2 = list2, list1
similar = True
for i in xrange(6):
    if list1[i] is not None and list1[i] != list2[i]:
        similar = False
        break

print(similar)
# False

但它们是相同的两个列表，应该仍然被认为是相似的。 IMO此代码检查两个列表中的None是正确的：

similar = True
for i in xrange(6):
    if ((list1[i] is not None and list2[i] is not None) and
           list1[i] != list2[i]):
        similar = False
        break

print(similar)
# True

Python：检查两个字符串列表是否“相似”的最快方法

2 个答案: