In Python how to obtain a partial view of a dict?

From the documentation:

CPython implementation detail: Keys and values are listed in an
arbitrary order which is non-random, varies across Python
implementations, and depends on the dictionary’s history of insertions
and deletions.

I’ve only toyed around at best with other Python implementations (eg PyPy, IronPython, etc), so I don’t know for certain if this is the case in all Python implementations, but the general idea of a dict/hashmap/hash/etc is that the keys are unordered.

That being said, you can use an OrderedDict from the collections library. OrderedDicts remember the order of the keys as you entered them.

If keys are someway sortable, you can do this:

head = dict([(key, myDict[key]) for key in sorted(myDict.keys())[:3]])

Or perhaps:

head = dict(sorted(mydict.items(), key=lambda: x:x[0])[:3])

Where x[0] is the key of each key/value pair.

import itertools 
def glance(d):
    return dict(itertools.islice(d.iteritems(), 3))

>>> x = {1:2, 3:4, 5:6, 7:8, 9:10, 11:12}
>>> glance(x)
{1: 2, 3: 4, 5: 6}

However:

>>> x['a'] = 2
>>> glance(x)
{1: 2, 3: 4, u'a': 2}

Notice that inserting a new element changed what the "first" three elements were in an unpredictable way. This is what people mean when they tell you dicts aren’t ordered. You can get three elements if you want, but you can’t know which three they’ll be.

I know this question is 3 years old but here a pythonic version (maybe simpler than the above methods) for Python 3.*:

[print(v) for i, v in enumerate(my_dict.items()) if i < n]

It will print the first n elements of the dictionary my_dict

For those who would rather solve this problem with pandas dataframes. Just stuff your dictionary mydict into a dataframe, rotate it, and get the first few rows:

pd.DataFrame(mydict, index=[0]).T.head()


0 hi0
1 hi1
2 hi2
3 hi3
4 hi4

one-up-ing @Neb’s solution with Python 3 dict comprehension:

{k: v for i, (k, v) in enumerate(my_dict.items()) if i < n}

It returns a dict rather than printouts

Edit:

in Python 3.x:
Without using libraries it’s possible to do it this way. Use method:

.items()

which returns a list of dictionary keys with values.

It’s necessary to convert it to a list otherwise an error will occur ‘my_dict’ object is not subscriptable. Then convert it to the dictionary. Now it’s ready to slice with square brackets.

dict(list(my_dict.items())[:3])

A quick and short solution can be this:

import pandas as pd

d = {"a": [1,2], "b": [2, 3], "c": [3, 4]}

pd.Series(d).head()

a    [1, 2]
b    [2, 3]
c    [3, 4]
dtype: object

list(reverse_word_index.items())[:10]

Change the number from 10 to however many items of the dictionary reverse_word_index you want to preview

This gives back a dictionary:

dict(list(my_dictname.items())[0:n])

If you just want to have a glance of your dict, then just do:

list(freqs.items())[0:n]

Order of items in a dictionary is preserved in Python 3.7+, so this question makes sense.

To get a dictionary with only 10 items from the start you can use pandas:

d = {"a": [1,2], "b": [2, 3], "c": [3, 4]}

import pandas as pd
result = pd.Series(d).head(10).to_dict()
print(result)

This will produce a new dictionary.

d = {"a": 1,"b": 2,"c": 3}
for i in list(d.items())[:2]:
     print('{}:{}'.format(d[i][0], d[i][1]))

a:1
b:2