通过索引替换python列表中的项目..失败？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/16916013/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Replacing an item in a python list by index.. failing?
提问by Jamus
Any idea why when I call:
知道为什么当我打电话时：
>>> hi = [1, 2]
>>> hi[1]=3
>>> print hi
[1, 3]
I can update a list item by its index, but when I call:
我可以通过索引更新列表项，但是当我调用时：
>>> phrase = "hello"
>>> for item in "123":
>>> list(phrase)[int(item)] = list(phrase)[int(item)].upper()
>>> print phrase
hello
It fails?
它失败？
Should be hELLo
应该 hELLo
回答by TerryA
You haven't initialised phrase(The listyou were intending to make) into a variable yet. So pretty much you have created a list in each loop, it being the exact same.
您尚未将phrase（list您打算制作的）初始化为变量。因此，您几乎在每个循环中都创建了一个列表，它完全相同。
If you were intending to actually change the characters of phrase, well that's not possible, as in python, strings are immutable.
如果您打算实际更改 的字符phrase，那是不可能的，因为在 python 中，字符串是不可变的。
Perhaps make phraselist = list(phrase), then edit the list in the for-loop. Also, you can use range():
也许 make phraselist = list(phrase)，然后在 for 循环中编辑列表。此外，您可以使用range()：
>>> phrase = "hello"
>>> phraselist = list(phrase)
>>> for i in range(1,4):
... phraselist[i] = phraselist[i].upper()
...
>>> print ''.join(phraselist)
hELLo
回答by Mike Müller
>>> phrase = "hello"
>>> list_phrase = list(phrase)
>>> for index in (1, 2, 3):
list_phrase[index] = phrase[index].upper()
>>> ''.join(list_phrase)
'hELLo'
If you prefer one-liner:
如果您更喜欢单线：
>>> ''.join(x.upper() if index in (1, 2, 3) else x for
index, x in enumerate(phrase))
'hELLo'
回答by oleg
consider that strings are immutable in python You can't modify existing string can create new.
考虑到python中的字符串是不可变的你不能修改现有的字符串可以创建新的。
''.join([c if i not in (1, 2, 3) else c.upper() for i, c in enumerate(phrase)])
''.join([c if i not in (1, 2, 3) else c.upper() for i, c in enumerate(phrase)])
回答by Inbar Rose
Another answer, just for fun :)
另一个答案，只是为了好玩:)
phrase = 'hello'
func = lambda x: x[1].upper() if str(x[0]) in '123' else x[1]
print ''.join(map(func, enumerate(phrase)))
# hELLo
To make this robust, I created a method: (because I am awesome, and bored)
为了使它健壮，我创建了一个方法：（因为我很棒，而且很无聊）
def method(phrase, indexes):
func = lambda x: x[1].upper() if str(x[0]) in indexes else x[1]
return ''.join(map(func, enumerate(phrase)))
print method('hello', '123')
# hELLo
回答by jfs
list()creates a newlist. Your loop creates and instantly discards two new lists on each iteration. You could write it as:
list()创建一个新列表。您的循环会在每次迭代中创建并立即丢弃两个新列表。你可以把它写成：
phrase = "hello"
L = list(phrase)
L[1:4] = phrase[1:4].upper()
print("".join(L))
Or without a list:
或者没有列表：
print("".join([phrase[:1], phrase[1:4].upper(), phrase[4:]]))
Strings are immutable in Python therefore to change it, you need to create a new string.
字符串在 Python 中是不可变的，因此要更改它，您需要创建一个新字符串。
Or if you are dealing with bytestrings, you could use bytearraywhich is mutable:
或者，如果您正在处理字节串，则可以使用bytearraywhich 是可变的：
phrase = bytearray(b"hello")
phrase[1:4] = phrase[1:4].upper()
print(phrase.decode())
If indexes are not consecutive; you could use an explicit for-loop:
如果索引不连续；您可以使用显式 for 循环：
indexes = [1, 2, 4]
for i in indexes:
L[i] = L[i].upper()
