Question

我有一个列表["a","b","c","a"]，其中我想用"a"替换["a_1","b_1","c_1"]，使列表变成["a_1","b_1","c_1","b","c", "a_1","b_1","c_1"]

在python中最快的方法是什么？

Answer 1

您可以对chain.from_iterable和三元语句使用理解：

from itertools import chain

L = list('abca')  # ['a', 'b', 'c', 'a']

rep_key, rep_val = ('a', ['a_1', 'b_1', 'c_1'])
res = list(chain.from_iterable([i] if i != rep_key else rep_val for i in L))

['a_1', 'b_1', 'c_1', 'b', 'c', 'a_1', 'b_1', 'c_1']

Answer 2

如果要变异您的列表，则可以依靠切片。实际上，这会替换您列表中的元素，而不创建一个新元素。

lst = ["a", "b", "c", "a"]
a_indices = [i for i, c in enumerate(lst) if c == "a"]

for i in reversed(a_indices):
    lst[i:i+1] = ['a_1', 'b_1', 'c_1']

print(lst) # ['a_1', 'b_1', 'c_1', 'b', 'c', 'a_1', 'b_1', 'c_1']

Answer 3

使用简单的for循环：

l = ["a", "b", "c", "a"]
for i in range(len(l)):
    if l[i] == "a":
        l[i] = "a_1"
        l+=["b_1", "c_1"]

>>> l
['a_1', 'b', 'c', 'a_1', 'b_1', 'c_1', 'b_1', 'c_1']
>>>

Answer 4

简单的列表理解应该可以解决问题。

abc = ["a","b","c","a"]

# Method 1 by using for loop
new_abc = []
for i in abc:
    if i == "a":
        new_abc += ["a_1","b_1","c_1"]
    else:
        new_abc.append(i)

# Method 2 by list comprehension
another_abc = [("a_1","b_1","c_1") if i == "a" else i for i in abc]
another_abc = [i for x in another_abc for i in x] #flatten the list

print (new_abc)
print (another_abc)

结果：

['a_1', 'b_1', 'c_1', 'b', 'c', 'a_1', 'b_1', 'c_1']
['a_1', 'b_1', 'c_1', 'b', 'c', 'a_1', 'b_1', 'c_1']

Answer 5

这是一个简单的生成器函数的好地方：

def lazy_replace(L, char, val):
    for item in L:
        if item == char:
            yield from val
        else:
            yield item

>>> L = list('abca')
>>> val = ['a_1', 'b_1', 'c_1']
>>> list(lazy_replace(L, 'a', val))
['a_1', 'b_1', 'c_1', 'b', 'c', 'a_1', 'b_1', 'c_1']

用python中的字符串列表替换一个字符串列表

5 个答案: