我可能是个笨蛋，但我会为它写一个单独的简单函数:

def convertElements( oldlist, convert_dict ):
  newlist = []
  for e in oldlist:
    if e in convert_dict:
      newlist.append(convert_dict[e])
    else:
      newlist.append(e)
  return newlist

然后根据需要调用它，如下所示:

a = [1,2,3,4,5,1,2,3,4,5,1]
a_new = convertElements(a, {1: 10})
## OUTPUT: a_new=[10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]

我可能是个笨蛋，但我会为它写一个单独的简单函数:

def convertElements( oldlist, convert_dict ):
  newlist = []
  for e in oldlist:
    if e in convert_dict:
      newlist.append(convert_dict[e])
    else:
      newlist.append(e)
  return newlist

然后根据需要调用它，如下所示:

a = [1,2,3,4,5,1,2,3,4,5,1]
a_new = convertElements(a, {1: 10})
## OUTPUT: a_new=[10, 2, 3, 4, 5, 10, 2, 3, 4, 5, 10]

尝试使用列表推导式和条件表达式。

>>> a=[1,2,3,1,3,2,1,1]
>>> [4 if x==1 else x for x in a]
[4, 2, 3, 4, 3, 2, 4, 4]

>>> a=[1,2,3,4,5,1,2,3,4,5,1]
>>> item_to_replace = 1
>>> replacement_value = 6
>>> indices_to_replace = [i for i,x in enumerate(a) if x==item_to_replace]
>>> indices_to_replace
[0, 5, 10]
>>> for i in indices_to_replace:
...     a[i] = replacement_value
... 
>>> a
[6, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6]
>>> 

如果你有几个值需要替换，你也可以使用字典:

a = [1, 2, 3, 4, 1, 5, 3, 2, 6, 1, 1]
replacements = {1:10, 2:20, 3:'foo'}
replacer = replacements.get  # For faster gets.

print([replacer(n, n) for n in a])

> [10, 20, 'foo', 4, 10, 5, 'foo', 20, 6, 10, 10]

注意，这种方法仅在要替换的元素是可哈希的情况下才有效。这是因为字典键必须是可哈希的。

在长列表和很少出现的情况下，使用list.index()比其他答案中给出的单步迭代方法快3倍左右。

def list_replace(lst, old=1, new=10):
    """replace list elements (inplace)"""
    i = -1
    try:
        while True:
            i = lst.index(old, i + 1)
            lst[i] = new
    except ValueError:
        pass