Login

partition template filters

Author:
SmileyChris
Posted:
September 2, 2007
Language:
Python
Version:
.96
Score:
7 (after 9 ratings)

A couple of useful template filters for splitting a list (or QuerySet) up into rows or columns.

  1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
"""
Template filters to partition lists into rows or columns.

A common use-case is for splitting a list into a table with columns::

    {% load partition %}
    <table>
    {% for row in mylist|columns:3 %}
        <tr>
        {% for item in row %}
            <td>{{ item }}</td>
        {% endfor %}
        </tr>
    {% endfor %}
    </table>
"""

from django.template import Library

register = Library()

def rows(thelist, n):
    """
    Break a list into ``n`` rows, filling up each row to the maximum equal
    length possible. For example::

        >>> l = range(10)

        >>> rows(l, 2)
        [[0, 1, 2, 3, 4], [5, 6, 7, 8, 9]]

        >>> rows(l, 3)
        [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9]]

        >>> rows(l, 4)
        [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

        >>> rows(l, 5)
        [[0, 1], [2, 3], [4, 5], [6, 7], [8, 9]]

        >>> rows(l, 9)
        [[0, 1], [2, 3], [4, 5], [6, 7], [8, 9], [], [], [], []]

        # This filter will always return `n` rows, even if some are empty:
        >>> rows(range(2), 3)
        [[0], [1], []]
    """
    try:
        n = int(n)
        thelist = list(thelist)
    except (ValueError, TypeError):
        return [thelist]
    list_len = len(thelist)
    split = list_len // n

    if list_len % n != 0:
        split += 1
    return [thelist[split*i:split*(i+1)] for i in range(n)]

def rows_distributed(thelist, n):
    """
    Break a list into ``n`` rows, distributing columns as evenly as possible
    across the rows. For example::

        >>> l = range(10)

        >>> rows_distributed(l, 2)
        [[0, 1, 2, 3, 4], [5, 6, 7, 8, 9]]

        >>> rows_distributed(l, 3)
        [[0, 1, 2, 3], [4, 5, 6], [7, 8, 9]]

        >>> rows_distributed(l, 4)
        [[0, 1, 2], [3, 4, 5], [6, 7], [8, 9]]

        >>> rows_distributed(l, 5)
        [[0, 1], [2, 3], [4, 5], [6, 7], [8, 9]]

        >>> rows_distributed(l, 9)
        [[0, 1], [2], [3], [4], [5], [6], [7], [8], [9]]

        # This filter will always return `n` rows, even if some are empty:
        >>> rows(range(2), 3)
        [[0], [1], []]
    """
    try:
        n = int(n)
        thelist = list(thelist)
    except (ValueError, TypeError):
        return [thelist]
    list_len = len(thelist)
    split = list_len // n

    remainder = list_len % n
    offset = 0
    rows = []
    for i in range(n):
        if remainder:
            start, end = (split+1)*i, (split+1)*(i+1)
        else:
            start, end = split*i+offset, split*(i+1)+offset
        rows.append(thelist[start:end])
        if remainder:
            remainder -= 1
            offset += 1
    return rows

def columns(thelist, n):
    """
    Break a list into ``n`` columns, filling up each column to the maximum equal
    length possible. For example::

        >>> from pprint import pprint
        >>> for i in range(7, 11):
        ...     print '%sx%s:' % (i, 3)
        ...     pprint(columns(range(i), 3), width=20)
        7x3:
        [[0, 3, 6],
         [1, 4],
         [2, 5]]
        8x3:
        [[0, 3, 6],
         [1, 4, 7],
         [2, 5]]
        9x3:
        [[0, 3, 6],
         [1, 4, 7],
         [2, 5, 8]]
        10x3:
        [[0, 4, 8],
         [1, 5, 9],
         [2, 6],
         [3, 7]]

        # Note that this filter does not guarantee that `n` columns will be
        # present:
        >>> pprint(columns(range(4), 3), width=10)
        [[0, 2],
         [1, 3]]
    """
    try:
        n = int(n)
        thelist = list(thelist)
    except (ValueError, TypeError):
        return [thelist]
    list_len = len(thelist)
    split = list_len // n
    if list_len % n != 0:
        split += 1
    return [thelist[i::split] for i in range(split)]

register.filter(rows)
register.filter(rows_distributed)
register.filter(columns)

def _test():
    import doctest
    doctest.testmod()

if __name__ == "__main__":
    _test()

More like this

  1. Template tag - list punctuation for a list of items by shapiromatron 2 months, 2 weeks ago
  2. JSONRequestMiddleware adds a .json() method to your HttpRequests by cdcarter 2 months, 3 weeks ago
  3. Serializer factory with Django Rest Framework by julio 9 months, 2 weeks ago
  4. Image compression before saving the new model / work with JPG, PNG by Schleidens 10 months, 1 week ago
  5. Help text hyperlinks by sa2812 11 months ago

Comments

SmileyChris (on December 30, 2007):

Actually, I notice from the comments that snippet 6 has a number of problems.

I have updated this snippet to fix all the problems noted there (complete with doctests ;))

#

CarlosPero (on December 31, 2008):

To have n number of columns with items in order across the rows, I added a "columns_across" function copied from "columns" but returns this instead:

return [thelist[n*i:n*(i+1)] for i in range(split)]

#

lgiordani (on May 29, 2012):

What about partitioning with a fixed lenght?

@register.filter
def partition_len(thelist, n):
    """
    Break a list into pieces with a maxmimum specified length.
    The last list may be shorter than the rest if the list doesn't
    break cleanly. That is::

        >>> l = range(10)

        >>> partition(l, 2)
        [[0, 1], [2, 3], [4, 5], [6, 7], [8, 9]]

        >>> partition(l, 3)
        [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

    """
    try:
        n = int(n)
        thelist = list(thelist)
    except (ValueError, TypeError):
        return [thelist]
    return (thelist[i:i+n] for i in xrange(0, len(thelist), n))

#

Please login first before commenting.