Question

我试图删除列表中包含重复字母的所有单词（以＆＃39; d＆＃39;＆＃39;＆＃39; r＆＃39;进行测试），但无法检查所有我想要的字母它来。

a = ["word","worrd","worrrd","wordd"]
alpha = ['d','r']
i = 0
x = 0
while x in range(0, len(alpha)):
    while i in range(0, len(a)):
        if a[i].count(alpha[x]) > 1:
            del(a[i])
            i = i - 1
        else:
            i = i + 1
    x = x + 1
print(a)

Answer 1

使用re.search()函数的简短解决方案：

import re

a = ["word","worrd","worrrd","wordd"]
result = [w for w in a if not re.search(r'([dr]).*\1', w)]

print(result)

输出：

[word]

([dr]).*\1 - 正则表达式模式，用于检查字符类[dr]中的某些字母是否在单词中出现至少两次

.* - 匹配0个或更多字符

\1 - 指向第一个捕获的群组(...)

另一个简短的替代方案是使用内置的all()函数：

a = ["word","worrd","worrrd","wordd"]
alpha = ['d','r']
result = [w for w in a if all(w.count(c) == 1 for c in alpha)]

print(result)

Answer 2

这是你的方法，只是我在第一次循环后添加了i = 0

a = ["word","worrd","worrrd","wordd"]
alpha = ['d','r']
i = 0
x = 0
while x in range(0, len(alpha)):
    i = 0
    while i in range(0, len(a)):
        if a[i].count(alpha[x]) > 1:
            del(a[i])
            i = i - 1
        else:
            i = i + 1
    x = x + 1
print(a)

['word']

Answer 3

你做得对，但你忘了重置＆＃34; i＆＃34;在while循环中

a = ["word","worrd","worrrd","wordd"]
alpha = ['d','r']
x = 0
while x in range(0, len(alpha)):
    # i should be reset here
    i = 0
    while i in range(0, len(a)):
        if a[i].count(alpha[x]) > 1:
            del(a[i])
            i = i - 1
        else:
            i = i + 1
    x = x + 1
print(a)

Answer 4

你可能会追求的是，当你连续两个d：s或r：s时。然后你想要一个正则表达式：

import re

a = ["word","worrd","worrrd","wordd", "worded"]
result = [w for w in a if not re.search(r'(dd)|(rr)', w)

print(result)

python：删除列表中包含重复字母的所有单词

4 个答案: