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
– Jerzyk, Commented Jun 14, 2016 at 10:09

Tom Dalton · Accepted Answer · 2016-06-15 09:52:31Z

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, 2016 at 9:52

answered Jun 14, 2016 at 10:12

Tom Dalton

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3 Comments

pythad Over a year ago

Counter object has no attribute iteritems

Sebastian Hoffmann Over a year ago

@pythad this is python 2 not 3. Use items() in 3

Tom Dalton Over a year ago

Updated answer to use items as that will work in both 2 and 3. Thanks for the comment.

Aswin Murugesh · Accepted Answer · 2016-06-14 10:11:49Z

3

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)")

answered Jun 14, 2016 at 10:11

Aswin Murugesh

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Comments

Padraic Cunningham · Accepted Answer · 2016-06-14 10:35:42Z

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.

sumit · Accepted Answer · 2016-06-14 10:13:06Z

2

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 !

answered Jun 14, 2016 at 10:13

sumit

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

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, 2020 at 9:12

CommunityBot

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answered Jun 14, 2016 at 10:12

Maximilian Peters

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5 Comments

Padraic Cunningham Over a year ago

This is O(n^2) as opposed to O(n) using a Counter

Maximilian Peters Over a year ago

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.

LarsVegas Over a year ago

this also creates a new set for every iteration.

Padraic Cunningham Over a year ago

At the time of my comment this had two more votes than the Counter approach so for future readers it is worth pointing out

Maximilian Peters Over a year ago

@Padraic Cunningham: Your approach is certainly more flexible and faster, +1.

miradulo · Accepted Answer · 2016-06-14 10:12:09Z

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, 2016 at 10:12

miradulo

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answered Jun 14, 2016 at 10:09

Sebastian Hoffmann

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Comments

Rafa · Accepted Answer · 2016-06-14 10:28:40Z

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, 2016 at 10:28

answered Jun 14, 2016 at 10:17

Rafa

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