python：删除列表中具有重复字母的所有单词

我正试图从包含重复字母的列表中删除所有单词（以“d”和“r”开头进行测试），但无法让它检查我希望它检查的所有字母

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)

---

这只是您的方法，除了我刚刚在第一个while循环之后添加了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']

使用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)

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)

输出：

[word]

---

（[dr]）.*\1

-用于检查字符类

[dr]

中的某个字母在一个单词中是否至少出现两次的正则表达式模式

*

-匹配0个或更多字符

\1

-指向第一个捕获的组

（…）

---

另一个简短的替代方案是使用内置功能：

import re

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

print(result)

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)

---

你做得对，但是你忘了在x的while循环中重置“i”的值

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)

---

你可能追求的是，当你有两个d:s或r:s在一排。那么您需要一个正则表达式，如下所示：

import re

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

print(result)

---

这也会错误地标记单词，比如

香蕉

@l3via，而他只想检查d和r。那么，

骑手

。用d和r重复字母，而不是两者都重复。这就是我在看到他的程序和他的问题后能够解释的。我想保留我的方法，但它只检查alpha的第一个字符list@EllisRourke，检查我的替代方法您没有在“x”循环中重置i，因此，这会导致下一个字符的匹配出现问题。谢谢，不知道我是如何错过的。：）