如何“更新”或“覆盖”python列表
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How to 'update' or 'overwrite' a python list
提问by pyLearner
aList = [123, 'xyz', 'zara', 'abc']
aList.append(2014)
print aList
which produces o/p [123, 'xyz', 'zara', 'abc', 2014]
产生 o/p [123, 'xyz', 'zara', 'abc', 2014]
What should be done to overwrite/update this list. I want the o/p to be
应该怎么做来覆盖/更新这个列表。我希望 o/p 是
[2014, 'xyz', 'zara', 'abc']
[2014, 'xyz', 'zara', 'abc']
采纳答案by Rahul Tripathi
You may try this
你可以试试这个
alist[0] = 2014
but if you are not sure about the position of 123 then you may try like this:
但如果您不确定 123 的位置，那么您可以尝试这样：
for idx, item in enumerate(alist):
if 123 in item:
alist[idx] = 2014
回答by Kerby82
What about replace the item if you know the position:
如果您知道位置，那么更换物品怎么样：
aList[0]=2014
Or if you don't know the position loop in the list, find the item and then replace it
或者如果你不知道列表中的位置循环，找到该项目然后替换它
aList = [123, 'xyz', 'zara', 'abc']
for i,item in enumerate(aList):
if item==123:
aList[i]=2014
print aList
回答by Ilya Kaznacheev
I think it is more pythonic:
我认为它更像pythonic：
aList.remove(123)
aList.insert(0, 2014)
more useful:
更有用：
def shuffle(list, to_delete, to_shuffle, index):
list.remove(to_delete)
list.insert(index, to_shuffle)
return
list = ['a', 'b']
shuffle(list, 'a', 'c', 0)
print list
>> ['c', 'b']
回答by Gabriel_Ferreira
I'm learning to code and I found this same problem. I believe the easier way to solve this is literaly overwriting the list like @kerby82 said:
我正在学习编码，我发现了同样的问题。我相信解决这个问题的更简单的方法是像@kerby82 所说的那样覆盖列表：
An item in a list in Python can be set to a value using the form
Python 中列表中的项目可以使用以下形式设置为一个值
x[n] = v
x[n] = v
Where xis the name of the list, nis the index in the array and vis the value you want to set.
其中x是列表的名称，n是数组中的索引，v是您要设置的值。
In your exemple:
在你的例子中：
aList = [123, 'xyz', 'zara', 'abc']
aList[0] = 2014
print aList
>>[2014, 'xyz', 'zara', 'abc']
回答by Haris Np
If you are trying to take a value from the same array and trying to update it, you can use the following code.
如果您尝试从同一个数组中获取一个值并尝试更新它，您可以使用以下代码。
{ 'condition': {
'ts': [ '5a81625ba0ff65023c729022',
'5a8161ada0ff65023c728f51',
'5a815fb4a0ff65023c728dcd']}
If the collection is userData['condition']['ts'] and we need to
如果集合是 userData['condition']['ts'] 并且我们需要
for i,supplier in enumerate(userData['condition']['ts']):
supplier = ObjectId(supplier)
userData['condition']['ts'][i] = supplier
The output will be
输出将是
{'condition': { 'ts': [ ObjectId('5a81625ba0ff65023c729022'),
ObjectId('5a8161ada0ff65023c728f51'),
ObjectId('5a815fb4a0ff65023c728dcd')]}
回答by innovacode
I would prefer it without enumerate and instead use "range" like this:
我更喜欢不枚举，而是使用“范围”，如下所示：
for item in range(0, len(alist)):
if 123 in alist[item]:
alist[item] = 2014
For those who are new to python it might be more readable and a little smarter to recap.
对于那些不熟悉 python 的人来说，它可能更具可读性，而且回顾起来也更聪明一些。
Regards P.
问候P。
