enumerate() for dictionary in python

Question

I know we use enumerate for iterating a list but I tried it in a dictionary and it didn't give an error.

CODE:

enumm = {0: 1, 1: 2, 2: 3, 4: 4, 5: 5, 6: 6, 7: 7}

for i, j in enumerate(enumm):
    print(i, j)

OUTPUT:

0 0

1 1

2 2

3 4

4 5

5 6

6 7

Can someone explain the output?

Answer 1

On top of the already provided answers there is a very nice pattern in Python that allows you to enumerate both keys and values of a dictionary.

The normal case you enumerate the keys of the dictionary:

example_dict = {1:'a', 2:'b', 3:'c', 4:'d'}

for i, k in enumerate(example_dict):
    print(i, k)

Which outputs:

0 1
1 2
2 3
3 4

But if you want to enumerate through both keys and values this is the way:

for i, (k, v) in enumerate(example_dict.items()):
    print(i, k, v)

Which outputs:

0 1 a
1 2 b
2 3 c
3 4 d

Answer 2

The first column of output is the index of each item in enumm and the second one is its keys. If you want to iterate your dictionary then use .items():

for k, v in enumm.items():
    print(k, v)

And the output should look like:

0 1
1 2
2 3
4 4 
5 5
6 6
7 7

Answer 3

Just thought I'd add, if you'd like to enumerate over the index, key, and values of a dictionary, your for loop should look like this:

for index, (key, value) in enumerate(your_dict.items()):
    print(index, key, value)

Answer 4

dict1={'a':1, 'b':'banana'}

To list the dictionary in Python 2.x:

for k,v in dict1.iteritems():
        print k,v 

In Python 3.x use:

for k,v in dict1.items():
        print(k,v)
# a 1
# b banana

Finally, as others have indicated, if you want a running index, you can have that too:

for i  in enumerate(dict1.items()):
   print(i)  

 # (0, ('a', 1))
 # (1, ('b', 'banana'))

But this defeats the purpose of a dictionary (map, associative array) , which is an efficient data structure for telephone-book-style look-up. Dictionary ordering could be incidental to the implementation and should not be relied upon. If you need the order, use OrderedDict instead.

Answer 5

Since you are using enumerate hence your i is actually the index of the key rather than the key itself.

So, you are getting 3 in the first column of the row 3 4 even though there is no key 3 .

enumerate iterates through a data structure(be it list or a dictionary) while also providing the current iteration number.

Hence, the columns here are the iteration number followed by the key in dictionary enum

Others Solutions have already shown how to iterate over key and value pair so I won't repeat the same in mine.

Answer 6

enumerate() when working on list actually gives the index and the value of the items inside the list. For example:

l = [1, 2, 3, 4, 5, 6, 7, 8, 9]
for i, j in enumerate(list):
    print(i, j)

gives

0 1
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9

where the first column denotes the index of the item and 2nd column denotes the items itself.

In a dictionary

enumm = {0: 1, 1: 2, 2: 3, 4: 4, 5: 5, 6: 6, 7: 7}
for i, j in enumerate(enumm):
    print(i, j)

it gives the output

0 0
1 1
2 2
3 4
4 5
5 6
6 7

where the first column gives the index of the key:value pairs and the second column denotes the keys of the dictionary enumm .

So if you want the first column to be the keys and second columns as values , better try out dict.iteritems() (Python 2) or dict.items() (Python 3)

for i, j in enumm.items():
    print(i, j)

output

0 1
1 2
2 3
4 4
5 5
6 6
7 7

Voila

Answer 7

That sure must seem confusing. So this is what is going on. The first value of enumerate (in this case i) returns the next index value starting at 0 so 0, 1, 2, 3, ... It will always return these numbers regardless of what is in the dictionary. The second value of enumerate (in this case j) is returning the values in your dictionary/enumm (we call it a dictionary in Python). What you really want to do is what roadrunner66 responded with.

Answer 8

1. Iterating over a Python dict means to iterate over its keys exactly the same way as with dict.keys()
2. The order of the keys is determined by the implementation code and you cannot expect some specific order:

   Keys and values are iterated over in an arbitrary order which is non-random, varies across Python implementations, and depends on the dictionary's history of insertions and deletions. If keys, values and items views are iterated over with no intervening modifications to the dictionary, the order of items will directly correspond.

That's why you see the indices 0 to 7 in the first column. They are produced by enumerate and are always in the correct order. Further you see the dict's keys 0 to 7 in the second column. They are not sorted.

Answer 9

Python3:

One solution:

enumm = {0: 1, 1: 2, 2: 3, 4: 4, 5: 5, 6: 6, 7: 7}
for i, k in enumerate(enumm):
    print("{}) d.key={}, d.value={}".format(i, k, enumm[k]))

---

Output:
0) enumm.key=0, enumm.value=1
1) enumm.key=1, enumm.value=2
2) enumm.key=2, enumm.value=3
3) enumm.key=4, enumm.value=4
4) enumm.key=5, enumm.value=5
5) enumm.key=6, enumm.value=6
6) enumm.key=7, enumm.value=7

An another example:

d = {1 : {'a': 1, 'b' : 2, 'c' : 3},
     2 : {'a': 10, 'b' : 20, 'c' : 30}
    }    
for i, k in enumerate(d):
        print("{}) key={}, value={}".format(i, k, d[k])

---

Output:    
    0) key=1, value={'a': 1, 'b': 2, 'c': 3}
    1) key=2, value={'a': 10, 'b': 20, 'c': 30}

Answer 10

You may find it useful to include index inside key:

d = {'a': 1, 'b': 2}
d = {(i, k): v for i, (k, v) in enumerate(d.items())}

Output:

{(0, 'a'): True, (1, 'b'): False}

Answer 11

d = {0: 'zero', '0': 'ZERO', 1: 'one', '1': 'ONE'}

print("List of enumerated d= ", list(enumerate(d.items())))

output:

List of enumerated d=  [(0, (0, 'zero')), (1, ('0', 'ZERO')), (2, (1, 'one')), (3, ('1', 'ONE'))]