從列表中刪除當前元素

Question

我有這個列表。 keys = ['Messi'，'Neymar'，'Xavi'，'Iniesta']我想遍歷列表以從列表中刪除當前元素。對於上面的列表，我想有類似的東西

Messi is out 
['Neymar', 'Xavi', 'Iniesta'] 
Neymar is out 
['Messi', 'Xavi', 'Iniesta'] 
Xavi is out 
['Messi', 'Neymar', 'Iniesta'] 
Iniesta is out 
['Messi', 'Neymar', 'Xavi'] 

這是我到目前爲止的代碼。它似乎並沒有工作

keys = ['Messi', 'Neymar', 'Xavi', 'Iniesta'] 
tmp_keys = keys 
length = len(keys) 
for player in keys: 
    if player in tmp_keys: 
     print player + " is out" 
     print tmp_keys 
     tmp_keys.remove(player) 
     tmp_keys = keys 

任何幫助將不勝感激。

Answer 1

您的代碼似乎是打印tmp_keys刪除的項目之前。我認爲，如果你轉換這些陳述的話，它會起作用。

一種更好的方式來複制列表可能僅僅是：

tmp_keys = list(keys) 

的話說tmp_keys = keys只是把tmp_keys另一個標記爲相同的列表爲keys

例如：

>>> a = [1,2,3] 
>>> b = a 

>>> print a  
[1,2,3] 

>>> print b 
[1,2,3] 

>>> b.append(4) 

>>> print a 
[1,2,3,4] 

>>> print b 
[1,2,3,4] 

你也可以這樣做：

from itertools import combinations 

keys = ['Messi', 'Neymar', 'Xavi', 'Iniesta'] 

c = combinations(keys, 3) 

>>> for i in c: 
...  print i 
('Messi', 'Neymar', 'Xavi') 
('Messi', 'Neymar', 'Iniesta') 
('Messi', 'Xavi', 'Iniesta') 
('Neymar', 'Xavi', 'Iniesta') 

for comb in c: 
    for name in keys: 
    if name not in comb: 
     print "{0} is out".format(name) 
     print list(comb) # without list() you will just get tuples 

輸出：

Iniesta is out 
['Messi', 'Neymar', 'Xavi'] 
Xavi is out 
['Messi', 'Neymar', 'Iniesta'] 
Neymar is out 
['Messi', 'Xavi', 'Iniesta'] 
Messi is out 
['Neymar', 'Xavi', 'Iniesta'] 

Answer 2

filter命令：

filter(lambda x: x!='Messi', ['Messi', 'Neymar', 'Xavi', 'Iniesta']) 
['Neymar', 'Xavi', 'Iniesta'] 

Answer 3

你顯然打印清單，你刪除它之前。您想在刪除它後打印它。你也應該複製的列表，而不是使用=在Python

keys = ['Messi', 'Neymar', 'Xavi', 'Iniesta'] 
tmp_keys = keys[:] 
length = len(keys) 
for player in keys: 
    if player in tmp_keys: 
    print player + " is out" 
    tmp_keys.remove(player) 
    print tmp_keys 
    tmp_keys = keys[:] 

http://docs.python.org/2/library/copy.html

賦值語句不復制的對象，他們創造的目標和對象之間的綁定。對於可變項目或包含可變項目的集合，有時需要副本，以便可以更改一個副本而不更改其他副本。

由於列表中不具有複製（），需​​要不同的方法，以防止從也影響鍵tmp_keys變化

作爲一個例子

keys = ['a', 'b', 'c', 'd', 'e'] 
tmp_keys = keys 

del tmp_keys(3) 
print keys 

OUTPUT：ABCE

tmp_keys = keys[:] 
print tmp_keys 
print keys 

OUTPUT：

a b c e 
a b c d e