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

Question

Is it possible to get a partial view of a dict in Python analogous of pandas df.tail()/df.head(). Say you have a very long dict, and you just want to check some of the elements (the beginning, the end, etc) of the dict. Something like:

dict.head(3)  # To see the first 3 elements of the dictionary.

{[1,2], [2, 3], [3, 4]}

Thanks

"the beginning" and "the end" have no meaning with Python dictionaries because their order is arbitrary. — , Feb 24 '15 at 19:28
What do you mean by "first three elements"? A `dict` has no inherent sense of key ordering. — , Feb 24 '15 at 19:28
I meant, the first three elements that would come up if you call the dict on your console.....Sometimes you do that, and if the dict is very long, you get a ton of data that clutters all your console space. — hernanavella, Feb 24 '15 at 19:30
@hernanavella: But you can't control or predict what three elements those will be. If you mean you want to just show an arbitrary subset of the dict elements, you can do that, but you won't know ahead of time exactly which elements you'll be seeing. — BrenBarn, Feb 24 '15 at 19:34
@iCodez I get that, but what if you just want to take a quick look at how the dict looks. — hernanavella, Feb 24 '15 at 19:35
I want to know how this is a dictionary `{[1,2], [2, 3], [3, 4]}` — Jdeep, Jul 13 '20 at 16:42
This is not a strange desire at all — I often have the situation where some calculation has returned a dictionary and I want to look inside it to make sure that it has the correct data and I haven't made a mistake. The order doesn't matter, I just want to see a sample of the data. This doesn't seem like it should be hard. I think the best answer is `list(my_dict.items())[:5]` — larapsodia, Nov 30 '22 at 18:33

score 63 · Accepted Answer · edited Mar 18 '22 at 16:20

63

Kinda strange desire, but you can get that by using this

from itertools import islice

# Python 2.x
dict(islice(mydict.iteritems(), 0, 2))

# Python 3.x
dict(islice(mydict.items(), 0, 2))

or for short dictionaries

# Python 2.x
dict(mydict.items()[0:2])

# Python 3.x
dict(list(mydict.items())[0:2])

edited Mar 18 '22 at 16:20

Arpad Horvath -- Слава Україні

1,744
1
17
39

answered Feb 24 '15 at 19:36

user4600699

1,465
11
10

1

perhaps `dict(sorted(mydict.items()[:3])` – heltonbiker Feb 24 '15 at 19:38
3

well if the order does matter, yes. But as far as i understand author don't need that. He just wants elements that would be printed at the beginning when printing dictionary and that is exactly what it is – user4600699 Feb 24 '15 at 19:41
2

`.items()` is an iterator, convert it to a list first, or it ll raise an error when sliced. – user Feb 24 '15 at 19:41
@user5061: No, OP specified Python 2.7. – John Y Feb 24 '15 at 19:43
1

# user 5061 In python2 items returns list. In python3 you can use islice to slice your iterator. – user4600699 Feb 24 '15 at 19:44
Not so strange, if you are using Jupiter notebook, a long dictionary will be shown completely, or rather it could crash the session. There is a missing functionality there, like for data frames, it only shows the top and bottom for longer data frames. – DISC-O Jun 03 '23 at 12:44

Michal · Answer 2 · 2021-01-15T07:04:43.350

17

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])

edited Jan 15 '21 at 07:04

answered Jun 21 '19 at 08:52

Michal

171
1
4

Could you give some context to your answer so it is useful for others? – Luis Sieira Jun 21 '19 at 09:01
2

Is it more understandable now? – Michal Jun 21 '19 at 11:43
works fine for my use-case where I need mock-data of a huge dict to do testing. – rambutan Sep 05 '22 at 13:37

score 8 · Answer 3 · edited Jul 10 '20 at 02:00

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.

Neb · Answer 4 · 2018-11-24T21:20:55.740

6

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

edited Nov 24 '18 at 21:20

answered Nov 14 '18 at 15:55

Neb

2,270
1
12
22

This only works if your keys are typically incrementing integers. What if you have negatives, large numbers, strings, or dates? – Wassadamo Nov 24 '18 at 19:30
1

@Wassadamo why would this not work with arbitrary keys? – Stefan_EOX Jun 21 '21 at 12:32
You're right. Enumerate indexes are used instead of the keys – Wassadamo Jun 21 '21 at 18:16

score 5 · Answer 5 · answered Feb 27 '19 at 07:03

5

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

answered Feb 27 '19 at 07:03

Little Bobby Tables

5,261
2
39
49

heltonbiker · Answer 6 · 2015-02-24T19:41:09.133

2

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.

edited Feb 24 '15 at 19:41

answered Feb 24 '15 at 19:35

heltonbiker

26,657
28
137
252

score 2 · Answer 7 · answered Nov 24 '18 at 19:39

2

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

answered Nov 24 '18 at 19:39

Wassadamo

1,176
12
32

score 1 · Answer 8 · answered Feb 24 '15 at 19:35

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.

He states in his comments that he already knows that. He just needs a small sample of dict contents. — user, Feb 24 '15 at 19:38

score 1 · Answer 9 · answered Nov 26 '20 at 11:33

1

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

answered Nov 26 '20 at 11:33

user566245

4,011
1
30
36

score 0 · Answer 10 · answered Jun 10 '20 at 22:35

0

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

answered Jun 10 '20 at 22:35

alejandro

521
8
18

Pay attention that this gives you an `object` instead of a `dict`. Meaning after `head = pd.Series(d).head()`, this will not work: `for name,group in head`... but this will: `for name in head.keys(): # \n # group = head[name]` – Cadoiz Jul 09 '20 at 19:32

score 0 · Answer 11 · answered May 19 '21 at 15:39

0

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]

answered May 19 '21 at 15:39

Tingmi

1

score 0 · Answer 12 · answered Sep 09 '21 at 10:38

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.

Deryc · Answer 13 · 2022-02-17T01:52:26.893

0

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

edited Feb 17 '22 at 01:52

answered Feb 06 '22 at 03:03

Deryc

58
5

1. needs `''` around `a, b, c`. 2. for first two elements, it's `[:2]`. 3. `i` is a tuple. 4. why not an f-string? 5. `i, d[i]` isn't right, you'd need `i[0], i[1]`. But `for k, v… print(f"{k}: {v}")` would be better. – Jack Deeth Feb 06 '22 at 13:41
(@JackDeeth about #4, question is flagged python 2.7, a f-string wouldn't be right.) Was this code tried ? As was already mentionned, .items() returns a tuple. You either want .keys() instead or need to change the print statement... – tgrandje Feb 18 '22 at 08:56

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

13 Answers13

Linked