Skip variable number of iterations in Python for loop

Question

I have a list and a for loop such as these:

mylist = ['foo','foo','foo','bar,'bar','hello']
for item in mylist:
    cp = mylist.count(item)
    print("You "+item+" are present in "+str(cp)+" copy(ies)")

Output:

You foo are present in 3 copy(ies)
You foo are present in 3 copy(ies)
You foo are present in 3 copy(ies)
You bar are present in 2 copy(ies)
You bar are present in 2 copy(ies)
You dude are present in 1 copy(ies)

Expected output:

You foo are present in 3 copy(ies)
You bar are present in 2 copy(ies)
You dude are present in 1 copy(ies)

The idea is thus to skip a variable number of iterations within the for loop, using something like this script (not working):

for item in mylist:
    cp = mylist.count(item)
    print("You "+item+" are present in "+str(cp)+" copy(ies)")
    continue(cp)

The script would thus "jump" cp elements in the for loop at every round and start doing again what it is asked at the item item + cp.

I know that you can use continue to skip multiple iterations (such as in this post) but I cannot figure out how to use continue to skip a variable number of iterations.

Thanks for your answer! :)

Edit: similar items are always next to each other.

is this list always sorted/grouped e.g. same elements are always together? — Jerzyk, Jun 14 '16 at 10:09

Tom Dalton · Accepted Answer · 2016-06-15T09:52:31.217

5

You could use a Counter:

from collections import Counter

mylist = ['foo','foo','foo','bar','bar','hello']
c = Counter(mylist)
for item, cp in c.items():
    print("You "+item+" are present in "+str(cp)+" copy(ies)")

edited Jun 15 '16 at 09:52

answered Jun 14 '16 at 10:12

Tom Dalton

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`Counter` object has no attribute `iteritems` – pythad Jun 14 '16 at 10:15
@pythad this is python 2 not 3. Use `items()` in 3 – Sebastian Hoffmann Jun 14 '16 at 10:46
Updated answer to use `items` as that will work in both 2 and 3. Thanks for the comment. – Tom Dalton Jun 15 '16 at 09:52

score 3 · Answer 2 · answered Jun 14 '16 at 10:11

You can use collections.Counter for your job:

>>> from collections import Counter
>>> Counter(['foo', 'foo', 'bar'])
Counter({'foo': 2, 'bar': 1})

Thus,

count_dict = Counter(mylist)
for item in count_dict:
    print("You "+item+" are present in "+str(count_dict[item[)+" copy(ies)")

Padraic Cunningham · Answer 3 · 2016-06-14T10:35:42.817

Since the elements are consecutive, you could use a groupby which would group the consecutive strings, it is just a matter of summing the length of each group to get the count:

from itertools import groupby
mylist = ['foo','foo','foo','bar','bar','hello']

for k,v in groupby(mylist):
    print("You {} are present in {} copy(ies)".format(k, sum(1 for _ in v)))

Output:

You foo are present in 3 copy(ies)
You bar are present in 2 copy(ies)
You hello are present in 1 copy(ies)

The most efficient way to generally get counts would be to use the dictionary logic like a Counter provided in the other answers, if you want to keep order you could use an OrderedDict to do the counting:

from collections import OrderedDict
mylist = ['foo','foo','foo','bar','bar','hello']
od = OrderedDict()
for ele in mylist:
    od.setdefault(ele, 0)
    od[ele] += 1

for tup in od.items():
    print("You {} are present in {} copy(ies)".format(*tup))

Which would output the same:

You foo are present in 3 copy(ies)
You bar are present in 2 copy(ies)
You hello are present in 1 copy(ies)

Both the groupby and the dict logic are O(n), using your list.count is quadratic.

score 2 · Answer 4 · edited Jun 20 '20 at 09:12

2

You can get the unique elements of a list using set

mylist = ['foo','foo','foo','bar','bar','hello']
uniq_list = set(my_list)
for item in uniq_list:
    cp = mylist.count(item)
    print("You "+item+" are present in "+str(cp)+" copy(ies)")

Output:

You bar are present in 2 copy(ies)

You hello are present in 1 copy(ies)

You foo are present in 3 copy(ies)

edited Jun 20 '20 at 09:12

Community

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1

answered Jun 14 '16 at 10:12

Maximilian Peters

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This is `O(n^2)` as opposed to `O(n)` using a Counter – Padraic Cunningham Jun 14 '16 at 10:23
I am aware of that but the question was not to find the fastest solution but how to do it. In case the list gets really long, a different approach would make certainly make sense. – Maximilian Peters Jun 14 '16 at 10:37
1

this also creates a new ``set`` for every iteration. – LarsVegas Jun 14 '16 at 10:39
1

At the time of my comment this had two more votes than the Counter approach so for future readers it is worth pointing out – Padraic Cunningham Jun 14 '16 at 10:40
@Padraic Cunningham: Your approach is certainly more flexible and faster, +1. – Maximilian Peters Jun 14 '16 at 10:43

score 2 · Answer 5 · answered Jun 14 '16 at 10:13

You can also do it this way

mylist = ['foo','foo','foo','bar','bar','hello']
prev = None
for item in mylist:
    if item != prev:
        cp = mylist.count(item)
        print("You "+item+" are present in "+str(cp)+" copy(ies)")
        prev = item

hope it may help !

score 1 · Answer 6 · edited Jun 14 '16 at 10:12

1

mylist = ['foo','foo','foo','bar','bar','hello']
last = None
for item in mylist:
    if item is last:
        continue

    last = item
    cp = mylist.count(item)
    print("You "+item+" are present in "+str(cp)+" copy(ies)")

This assumes that the list is ordered so that identical objects are next to each other.

edited Jun 14 '16 at 10:12

miradulo

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answered Jun 14 '16 at 10:09

Sebastian Hoffmann

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Rafa · Answer 7 · 2016-06-14T10:28:40.013

1

set and list counts would do the trick:

mylist = ['foo','foo','foo','bar','bar','hello']
for item in set(mylist):
    print("You "+item+" are present in "+str(mylist.count(item))+" copy(ies)")

output:

You foo are present in 3 copy(ies)
You bar are present in 2 copy(ies)
You hello are present in 1 copy(ies)

edited Jun 14 '16 at 10:28

answered Jun 14 '16 at 10:17

Rafa

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Skip variable number of iterations in Python for loop

7 Answers7