CODEWITHSUNDEEP
python
Mu Mind
Mu Mind

Reputation: 11204

Simple way to group items into buckets

I often want to bucket an unordered collection in python. itertools.groubpy does the right sort of thing but almost always requires massaging to sort the items first and catch the iterators before they're consumed.

Is there any quick way to get this behavior, either through a standard python module or a simple python idiom?

>>> bucket('thequickbrownfoxjumpsoverthelazydog', lambda x: x in 'aeiou')
{False: ['t', 'h', 'q', 'c', 'k', 'b', 'r', 'w', 'n', 'f', 'x', 'j', 'm', 'p',
    's', 'v', 'r', 't', 'h', 'l', 'z', 'y', 'd', 'g'],
 True: ['e', 'u', 'i', 'o', 'o', 'u', 'o', 'e', 'e', 'a', 'o']}
>>> bucket(xrange(21), lambda x: x % 10)
{0: [0, 10, 20],
 1: [1, 11],
 2: [2, 12],
 3: [3, 13],
 4: [4, 14],
 5: [5, 15],
 6: [6, 16],
 7: [7, 17],
 8: [8, 18],
 9: [9, 19]}

Upvotes: 19

Views: 37298

Answers (6)

Jelen
Jelen

Reputation: 633

You can use generator like this "yield" means, that function returns on every go through

array.pop cut last element from array

def get_bucket(arr, n):
    while len(arr) > 0:
        l = len(arr)
        bucket = []
        for i in range(n if n < l else l):
            bucket.append(arr.pop())
        yield bucket


for b in get_bucket(list(range(123)), 5):
    print(b)

[122, 121, 120, 119, 118]
[117, 116, 115, 114, 113]
[112, 111, 110, 109, 108]
[107, 106, 105, 104, 103]
[102, 101, 100, 99, 98]
[97, 96, 95, 94, 93]
[92, 91, 90, 89, 88]
[87, 86, 85, 84, 83]
[82, 81, 80, 79, 78]
[77, 76, 75, 74, 73]
[72, 71, 70, 69, 68]
[67, 66, 65, 64, 63]
[62, 61, 60, 59, 58]
[57, 56, 55, 54, 53]
[52, 51, 50, 49, 48]
[47, 46, 45, 44, 43]
[42, 41, 40, 39, 38]
[37, 36, 35, 34, 33]
[32, 31, 30, 29, 28]
[27, 26, 25, 24, 23]
[22, 21, 20, 19, 18]
[17, 16, 15, 14, 13]
[12, 11, 10, 9, 8]
[7, 6, 5, 4, 3]
[2, 1, 0]

Upvotes: 0

Boern
Boern

Reputation: 7752

If its a pandas.DataFrame the following also works, utilizing pd.cut()

from sklearn import datasets
import pandas as pd

# import some data to play with
iris = datasets.load_iris()
df_data = pd.DataFrame(iris.data[:,0])  # we'll just take the first feature

# bucketize
n_bins = 5
feature_name = iris.feature_names[0].replace(" ", "_")
my_labels = [str(feature_name) + "_" + str(num) for num in range(0,n_bins)]
pd.cut(df_data[0], bins=n_bins, labels=my_labels)

yielding

0      0_1
1      0_0
2      0_0
[...]

In case you don't set the labels, the output is going to like this

0       (5.02, 5.74]
1      (4.296, 5.02]
2      (4.296, 5.02]
[...]

Upvotes: 2

Thomas Perl
Thomas Perl

Reputation: 2348

Here's a variant of partition() from above when the predicate is boolean, avoiding the cost of a dict/defaultdict:

def boolpartition(seq, pred):
    passing, failing = [], []
    for item in seq:
        (passing if pred(item) else failing).append(item)
    return passing, failing

Example usage:

>>> even, odd = boolpartition([1, 2, 3, 4, 5], lambda x: x % 2 == 0)
>>> even
[2, 4]
>>> odd
[1, 3, 5]

Upvotes: 2

DSM
DSM

Reputation: 353359

This has come up several times before -- (1), (2), (3) -- and there's a partition recipe in the itertools recipes, but to my knowledge there's nothing in the standard library.. although I was surprised a few weeks ago by accumulate, so who knows what's lurking there these days? :^)

When I need this behaviour, I use

from collections import defaultdict

def partition(seq, key):
    d = defaultdict(list)
    for x in seq:
        d[key(x)].append(x)
    return d

and get on with my day.

Upvotes: 23

korylprince
korylprince

Reputation: 3009

Edit:

Using DSM's answer as a start, here is a slightly more concise, general answer:

d = defaultdict(list)
map(lambda x: d[x in 'aeiou'].append(x),'thequickbrownfoxjumpsoverthelazydog')

or

d = defaultdict(list)
map(lambda x: d[x %10].append(x),xrange(21))
#

Here is a two liner:

d = {False:[],True:[]}
filter(lambda x: d[True].append(x) if x in 'aeiou' else d[False].append(x),"thequickbrownfoxjumpedoverthelazydogs")

Which can of course be made a one-liner:

d = {False:[],True:[]};filter(lambda x: d[True].append(x) if x in 'aeiou' else d[False].append(x),"thequickbrownfoxjumpedoverthelazydogs")

Upvotes: -1

grieve
grieve

Reputation: 13508

Here is a simple two liner

d = {}
for x in "thequickbrownfoxjumpsoverthelazydog": d.setdefault(x in 'aeiou', []).append(x)

Edit:

Just adding your other case for completeness.

d={}
for x in xrange(21): d.setdefault(x%10, []).append(x)

Upvotes: 6

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