- 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()
|
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Comments
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 ;))
#
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)]
#
What about partitioning with a fixed lenght?
#
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