python 使用 for 循环删除列表中的项目
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Delete item in a list using a for-loop
提问by kame
I have an array with subjects and every subject has connected time. I want to compare every subjects in the list. If there are two of the same subjects, I want to add the times of both subjects, and also want to delete the second subject information (subject-name and time).
我有一个包含主题的数组，每个主题都有相关的时间。我想比较列表中的每个主题。如果有两个相同的科目，我想添加两个科目的时间，还想删除第二个科目信息（科目名称和时间）。
But If I delete the item, the list become shorter, and I get an out-of-range-error. I tried to make the list shorter with using subjectlegth-1, but this also don't work.
但是如果我删除该项目，列表会变短，并且会出现超出范围的错误。我试图通过使用 subjectlegth-1 来缩短列表，但这也不起作用。
...
subjectlegth = 8
for x in range(subjectlength):
for y in range(subjectlength):
if subject[x] == subject[y]:
if x != y:
#add
time[x] = time[x] + time[y]
#delete
del time[y]
del subject[y]
subjectlength = subjectlength - 1
回答by Chris Jester-Young
Iterate backwards, if you can:
如果可以，向后迭代：
for x in range(subjectlength - 1, -1, -1):
and similarly for y.
对于y.
回答by Ignacio Vazquez-Abrams
If the elements of subjectare hashable:
如果 的元素是subject可散列的：
finalinfo = {}
for s, t in zip(subject, time):
finalinfo[s] = finalinfo.get(s, 0) + t
This will result in a dict with subject: timekey-value pairs.
这将导致一个带有subject: time键值对的字典。
回答by Ned Batchelder
The best practice is to make a new list of the entries to delete, and to delete them after walking the list:
最佳做法是为要删除的条目创建一个新列表，并在遍历列表后删除它们：
to_del = []
subjectlength = 8
for x in range(subjectlength):
for y in range(x):
if subject[x] == subject[y]:
#add
time[x] = time[x] + time[y]
to_del.append(y)
to_del.reverse()
for d in to_del:
del subject[d]
del time[d]
回答by MAK
An alternate way would be to create the subject and time lists anew, using a dict to sum up the times of recurring subjects (I am assuming subjects are strings i.e. hashable).
另一种方法是重新创建主题和时间列表，使用 dict 来总结重复主题的时间（我假设主题是字符串，即可散列）。
subjects=['math','english','necromancy','philosophy','english','latin','physics','latin']
time=[1,2,3,4,5,6,7,8]
tuples=zip(subjects,time)
my_dict={}
for subject,t in tuples:
try:
my_dict[subject]+=t
except KeyError:
my_dict[subject]=t
subjects,time=my_dict.keys(), my_dict.values()
print subjects,time
回答by Reza Dodge
Though a whileloop is certainly a better choice for this, if you insist on using a forloop, one can replace the listelements-to-be-deleted with None, or any other distinguishable item, and redefine the listafter the forloop. The following code removes even elements from a list of integers:
虽然一while环无疑是这一个更好的选择，如果你坚持使用一个for回路，可以换list用无，或其他任何区分item元素将要删除的，并重新定义list后for循环。以下代码从整数列表中删除偶数元素：
nums = [1, 1, 5, 2, 10, 4, 4, 9, 3, 9]
for i in range(len(nums)):
# select the item that satisfies the condition
if nums[i] % 2 == 0:
# do_something_with_the(item)
nums[i] = None # Not needed anymore, so set it to None
# redefine the list and exclude the None items
nums = [item for item in nums if item is not None]
# num = [1, 1, 5, 9, 3, 9]
In the case of the question in this post:
对于这篇文章中的问题：
...
for i in range(subjectlength - 1):
for j in range(i+1, subjectlength):
if subject[i] == subject[j]:
#add
time[i] += time[j]
# set to None instead of delete
time[j] = None
subject[j] = None
time = [item for item in time if item is not None]
subject = [item for item in subject if item is not None]
