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How to iterate a dict with enumerate such that I could unpack the index, key and value at the time of iteration?

Something like:

for i, (k, v) in enumerate(mydict):
    # some stuff

I want to iterate through the keys and values in a dictionary called mydict and count them, so I know when I'm on the last pair.

cottontail
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themink
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2 Answers2

116

Instead of using mydict, you should be using mydict.items() with enumerate as:

for i, (k, v) in enumerate(mydict.items()):
    # your stuff

Sample example:

mydict = {1: 'a', 2: 'b'}
for i, (k, v) in enumerate(mydict.items()):
    print("index: {}, key: {}, value: {}".format(i, k, v))

# which will print:
# -----------------
# index: 0, key: 1, value: a
# index: 1, key: 2, value: b

Explanation:

  • enumerate() returns an iterator object which contains tuples in the format: [(index, list_element), ...]
  • dict.items() returns an iterator object (in Python 3.x. It returns list in Python 2.7) in the format: [(key, value), ...]
  • On combining together, enumerate(dict.items()) will return an iterator object containing tuples in the format: [(index, (key, value)), ...]
Moinuddin Quadri
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0

It's also possible to use itertools.count to count along.

from itertools import count
for i, (k, v) in zip(count(), mydict.items()):
    # do something

This is useful especially if you want the counter to be fractional numbers. For example,

mydict = dict(zip(range(3), range(10, 13)))

for i, (k, v) in zip(count(step=0.5), mydict.items()):
    print(f"index={i:<3}, key={k}, value={v}")
    
# index=0  , key=0, value=10
# index=0.5, key=1, value=11
# index=1.0, key=2, value=12
cottontail
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