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?
This question is related to
python
python-3.x
dictionary
enumerate
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)
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.
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'))]
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
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
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}
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
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.
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}
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.
dict.keys()
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.
Source: Stackoverflow.com