Python - 如何更改列表列表中的值？
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Python - How to change values in a list of lists?
提问by Wichid Nixin
I have a list of lists, each list within the list contains 5 items, how do I change the values of the items in the list? I have tried the following:
我有一个列表列表，列表中的每个列表包含 5 个项目，如何更改列表中项目的值？我尝试了以下方法：
for [itemnumber, ctype, x, y, delay] in execlist:
if itemnumber == mynumber:
ctype = myctype
x = myx
y = myy
delay = mydelay
Originally I had a list of tuples but I realized I cant change values in a tuple so I switched to lists but I still cant change any of the values. If I print ctype, x, y, delay, myctype, myx, myy, or mydelay from within the for loop it appears that everything is working but if I print execlist afterwards I see that nothing has changed.
最初我有一个元组列表，但我意识到我无法更改元组中的值，所以我切换到列表，但我仍然无法更改任何值。如果我从 for 循环中打印 ctype、x、y、delay、myctype、myx、myy 或 mydelay，似乎一切正常，但如果我之后打印 execlist，我会发现没有任何变化。
采纳答案by Jeff Silverman
The problem is that you are creating a copy of the list and then modifying the copy. What you want to do is modify the original list. Try this instead:
问题是您正在创建列表的副本，然后修改副本。您要做的是修改原始列表。试试这个：
for i in range(len(execlist)):
if execlist[i][0] == mynumber:
execlist[i][1] = myctype
execlist[i][2] = myx
execlist[i][3] = myy
execlist[i][4] = mydelay
回答by BenTrofatter
You need to assign via indexes. Let's say you've got a list of lists, where the inner lists each have 5 items like you describe. If you want to iterate through them and change the value of the second item in each inner list, you could do something like:
您需要通过索引进行分配。假设您有一个列表列表，其中每个内部列表都有您描述的 5 个项目。如果您想遍历它们并更改每个内部列表中第二项的值，您可以执行以下操作：
l = [[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14]]
for i in l:
i[1] = "spam"
print l
(output) [[0, "spam", 2, 3, 4], [5, "spam", 7, 8, 9], [10, "spam", 12, 13, 14]]
回答by RickyA
Variable unpacking does not seem to pass the reference but copies the values. An solution would be to do it like this:
变量解包似乎没有传递引用而是复制值。一个解决方案是这样做：
foo = [[1, "gggg"], [3, "zzzz"]]
for item in foo:
item[0] = 2
item[1] = "ffff"
print(foo)
>>>> [[2, 'ffff'], [2, 'ffff']]
回答by Tim Pietzcker
You could use enumerate():
你可以使用enumerate()：
for index, sublist in enumerate(execlist):
if sublist[0] == mynumber:
execlist[index][1] = myctype
execlist[index][2] = myx
execlist[index][3] = myy
execlist[index][4] = mydelay
# break
You can remove the #if execlistonly contains at most one sublist whose first item can equal mynumber; otherwise, you'll cycle uselessly through the entire rest of the list.
您可以删除#ifexeclist只包含最多一个其第一项可以等于的子列表mynumber；否则，您将无用地循环遍历列表的其余部分。
And if the itemnumbersare in fact unique, you might be better off with a dictionary or at least an OrderedDict, depending on what else you intend to do with your data.
如果itemnumbers实际上是唯一的，则最好使用字典或至少使用OrderedDict，这取决于您打算对数据做什么。
回答by ApproachingDarknessFish
Don't assign local variables in lists. In the loop
不要在列表中分配局部变量。在循环
for i in lis:
i = 5
Just sets the variable ito 5 and leaves the actual contents of the list unchanged. Instead, you have to assign it directly:
只需将变量设置i为 5 并保持列表的实际内容不变。相反，您必须直接分配它：
for i in range(len(lis)):
lis[i] = 5
The same applied for lists of lists, although in this case the local variable doesn't have to be assigned so you can use the for...inconstruct.
这同样适用于列表列表，尽管在这种情况下不必分配局部变量，因此您可以使用该for...in构造。
for i in listoflists:
for i2 in range(len(i)):
i[i2] = 5 #sets all items in all lists to 5
回答by martineau
Changing the variables assigned in the fordoes not change the list. Here's a fairly readable way to do what you want:
更改在 中分配的变量for不会更改列表。这是一个相当易读的方法来做你想做的事：
execlist = [
#itemnumber, ctype, x, y, delay
[1, 2, 3, 4, 5],
[6, 7, 8, 9, 10],
[11, 12, 13, 14, 15],
]
mynumber, myctype, myx, myy, mydelay = 6, 100, 101, 102, 104
for i, sublist in enumerate(execlist):
if sublist[0] == mynumber:
execlist[i] = [mynumber, myctype, myx, myy, mydelay]
print execlist
Output:
输出：
[[1, 2, 3, 4, 5], [6, 100, 101, 102, 104], [11, 12, 13, 14, 15]]
回答by dbrobins
The iterator variables are copies of the original (Python does not have a concept of a reference as such, although structures such as lists are referred to by reference). You need to do something like:
迭代器变量是原始变量的副本（Python 本身没有引用的概念，尽管列表等结构是通过引用引用的）。您需要执行以下操作：
for item in execlist:
if item[0] == mynumber:
item[1] = ctype
item[2] = myx
item[3] = myy
item[4] = mydelay
itemitself is a copy too, but it is a copy of a reference to the original nested list, so when you refer to its elements the original list is updated.
item本身也是一个副本，但它是对原始嵌套列表的引用的副本，因此当您引用其元素时，原始列表会更新。
This isn't as convenient since you don't have the names; perhaps a dictionary or class would be a more convenient structure.
这不太方便，因为您没有名字；也许字典或类会是一个更方便的结构。
回答by Erik Ware
Here is an example I used recently, it was a real mind bender but hopefully it can help out some one!
这是我最近使用的一个例子，它是一个真正的头脑弯曲，但希望它可以帮助一些人！
pivot_peaks is a list of pandas data frames some of them are duplicates.
pivot_peaks 是熊猫数据框的列表，其中一些是重复的。
# Create a new list with only a single entry for each item
new_list = []
for obj_index, obj in enumerate(pivot_peaks):
add_new_frame = False
if len(new_list) == 0:
new_list.append(obj)
else:
for item in new_list:
if item.equals(obj):
add_new_frame = False
break
else:
add_new_frame = True
if add_new_frame == True:
new_list.append(obj)
