从 python 元组中弹出/删除项目
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pop/remove items out of a python tuple
提问by Steve Grafton
I am not sure if I can make myself clear but will try.
我不确定我是否可以让自己清楚，但会尝试。
I have a tuple in python which I go through as follows (see code below). While going through it, I maintain a counter (let's call it 'n') and 'pop' items that meet a certain condition.
我在 python 中有一个元组，我按如下方式进行（见下面的代码）。在浏览它时，我维护了一个计数器（我们称之为“n”）和满足特定条件的“流行”项目。
Now of course once I pop the first item, the numbering all goes wrong, how can I do what I want to do more elegantly while removing only certain entries of a tuple on the fly?
现在当然，一旦我弹出第一个项目，编号就会出错，我怎样才能更优雅地做我想做的事情，同时只删除元组的某些条目？
for x in tupleX:
n=0
if (condition):
tupleX.pop(n)
n=n+1
采纳答案by Steve Grafton
ok I figured out a crude way of doing it.
好的，我想出了一个粗略的方法。
I store the "n" value in the for loop when condition is satisfied in a list (lets call it delList) then do the following:
当列表中满足条件时，我将“n”值存储在 for 循环中（我们称之为 delList），然后执行以下操作：
for ii in sorted(delList, reverse=True):
tupleX.pop(ii)
Any other suggestions are welcome too.
也欢迎任何其他建议。
回答by tk.
Maybe you want dictionaries?
也许你想要字典？
d = dict( (i,value) for i,value in enumerate(tple))
while d:
bla bla bla
del b[x]
回答by SingleNegationElimination
As DSMmentions, tuple's are immutable, but even for lists, a more elegant solution is to use filter:
如前所述DSM，tuple's 是不可变的，但即使对于列表，更优雅的解决方案是使用filter：
tupleX = filter(str.isdigit, tupleX)
or, if conditionis not a function, use a comprehension:
或者，如果condition不是函数，则使用推导式：
tupleX = [x for x in tupleX if x > 5]
if you really need tupleX to be a tuple, use a generator expression and pass that to tuple:
如果您确实需要 tupleX 作为元组，请使用生成器表达式并将其传递给tuple：
tupleX = tuple(x for x in tupleX if condition)
回答by Reza K Ghazi
There is a simple but practical solution.
有一个简单但实用的解决方案。
As DSM said, tuples are immutable, but we know Lists are mutable. So if you change a tuple to a list, it will be mutable. Then you can delete the items by the condition, then after changing the type to a tuple again. That's it.
正如 DSM 所说，元组是不可变的，但我们知道列表是可变的。因此，如果将元组更改为列表，它将是可变的。然后您可以按条件删除项目，然后再次将类型更改为元组。就是这样。
Please look at the codes below:
请看下面的代码：
tuplex = list(tuplex)
for x in tuplex:
if (condition):
tuplex.pop(tuplex.index(x))
tuplex = tuple(tuplex)
print(tuplex)
For example, the following procedure will delete all even numbers from a given tuple.
例如，以下过程将从给定的元组中删除所有偶数。
tuplex = (1, 2, 3, 4, 5, 6, 7, 8, 9)
tuplex = list(tuplex)
for x in tuplex:
if (x % 2 == 0):
tuplex.pop(tuplex.index(x))
tuplex = tuple(tuplex)
print(tuplex)
if you test the type of the last tuplex, you will find it is a tuple.
如果你测试最后一个tuplex的类型，你会发现它是一个tuple。
Finally, if you want to define an index counter as you did (i.e., n), you should initialize it before the loop, not in the loop.
最后，如果你想像你所做的那样定义一个索引计数器（即 n），你应该在循环之前初始化它，而不是在循环中。
回答by Abdulvakaf K
Yes we can do it. First convert the tuple into an list, then delete the element in the list after that again convert back into tuple.
是的，我们可以做到。首先将元组转换为列表，然后删除列表中的元素，然后再次转换回元组。
Demo:
演示：
my_tuple = (10, 20, 30, 40, 50)
# converting the tuple to the list
my_list = list(my_tuple)
print my_list # output: [10, 20, 30, 40, 50]
# Here i wanna delete second element "20"
my_list.pop(1) # output: [10, 30, 40, 50]
# As you aware that pop(1) indicates second position
# Here i wanna remove the element "50"
my_list.remove(50) # output: [10, 30, 40]
# again converting the my_list back to my_tuple
my_tuple = tuple(my_list)
print my_tuple # output: (10, 30, 40)
Thanks
谢谢
回答by ShadowPuppy
The best solution is the tuple applied to a list comprehension, but to extract one item this could work:
最好的解决方案是将元组应用于列表理解，但要提取一个项目，这可以工作：
def pop_tuple(tuple, n):
return tuple[:n]+tuple[n+1:], tuple[n]
def pop_tuple(tuple, n):
return tuple[:n]+tuple[n+1:], tuple[n]
