Delete item in a list using a for-loop

Question

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 ti

Answer 1

Iterate backwards, if you can:

for x in range(subjectlength - 1, -1, -1):

and similarly for y.

Answer 2

Though a while loop is certainly a better choice for this, if you insist on using a for loop, one can replace the list elements-to-be-deleted with None, or any other distinguishable item, and redefine the list after the for loop. The following code removes even elements from a list of integers:

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:

...
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]

Answer 3

The best practice is to make a new list of the entries to delete, and to delete them after walking the list:

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]

Answer 4

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).

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

Answer 5

If the elements of subject are hashable:

finalinfo = {}

for s, t in zip(subject, time):
  finalinfo[s] = finalinfo.get(s, 0) + t

This will result in a dict with subject: time key-value pairs.