Python中for.. in循环的意外输出

Question

我想运行一个 for 循环来检查字符串中的每个字母，如果 wanted 列表包含该字母，那么我希望该字母保留，如果 wanted 不包含该字母，那么我想删除它。

phrase = "Don't panic"
pList = list(phrase)
print(pList)
wanted = ["o","n","t","a","p"]
for anything in pList:
    print("Now checking", anything)
    if anything not in wanted:
        print("Removed",anything) 
        pList.remove(anything)
    elif anything in wanted:
        print("Didn't remove", anything)

print(pList)

以下代码返回此输出-

['D', 'o', 'n', "'", 't', ' ', 'p', 'a', 'n', 'i', 'c']                                                   

Now checking D                                                                                       
Removed D                                                                                                                                                                                           
Now checking n                                                                                          
Didn't remove n                                                                                         
Now checking '                                                                                           
Removed '                                                                                                                                                                                               
Now checking                                                                                              
Removed                                                                                                                                                                                                 
Now checking a                                                                                           
Didn't remove a                                                                                         
Now checking n                                                                                           
Didn't remove n                                                                                           
Now checking i                                                                                          
Removed i                                                                                                                                                                                       
['o', 'n', 't', 'p', 'a', 'n', 'c']   

代码没有删除最后一个字母（即“c”），字母“o”、“t”、“p”和“c”没有显示Now checking输出。

我尝试删除一些行然后就跑了 -

phrase = "Don't panic"
pList = list(phrase)
print(pList)
wanted = ["o","n","t","a","p"]
for anything in pList:
    print("Now checking", anything)

这一次的输出有意义 -

Now checking D                                                                                            
Now checking o                                                                                            
Now checking n                                                                                            
Now checking '                                                                                            
Now checking t                                                                                            
Now checking                                                                                              
Now checking p                                                                                            
Now checking a                                                                                            
Now checking n                                                                                            
Now checking i                                                                                            
Now checking c 

那么为什么在运行完整代码时不会发生这种情况？

Answer 1

您在迭代列表时正在对其进行编辑。正因为如此，大小会在中途改变，你最终会出现奇怪的行为。您可以改为创建一个新列表来保存过滤后的字符。

phrase = "Don't panic"
pList = list(phrase)
new_plist = []
wanted = ["o","n","t","a","p"]
for i, anything in enumerate(pList):
    print("Now checking", anything)
    if anything in wanted:
        print("Didn't remove", anything)
        new_plist.append(anything)
    else:
        print("Removed", anything)

print(new_plist)

使用列表推导的更简单的方法是

pList = ''.join([i for i in pList if i not in wanted])

Answer 2

其他答案和 cmets 都是 100% 正确的 - 迭代列表时切勿修改列表！可能有助于解决这类问题的方法是考虑逆向问题：不是从列表中删除，而是反转条件并将添加到一个新列表。通过这种方式，它基于输入而不是改变输入来构建新的输出 = 更少的副作用、更简洁的代码、函数式编程原则等。@coldspeed 的列表推导是一种优雅的解决方案。

对您的原始代码更改最少的替代解决方案：

phrase = "Don't panic"
pList = list(phrase)
print(pList)
wanted = ["o","n","t","a","p"]
finalList = []
for anything in pList:
    print("Now checking", anything)
    if anything in wanted:
        finalList.append(anything)
        print("Didn't remove", anything)
    else:
        print("Removed",anything)

print(finalList)

现在pList 不会被修改，新的finalList 包含所需的结果。 （编辑：意识到@coldspeed 的代码现在几乎和这个完全一样——我最初提交的时候不是这样。）