Various list utilities.
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 | #!/usr/bin/env python
#
# Some of these are adapted from other djangosnippets.
def has_consecutive_duplicates_of(seq, item):
"""Returns whether duplicate consecutive elements are in the sequence.
Args:
seq: The sequence.
item: Which item to test for duplicates.
>>> has_consecutive_duplicates_of([], 5)
False
>>> has_consecutive_duplicates_of([1, 2, 2, 3], 5)
False
>>> has_consecutive_duplicates_of([1, 2, 2, 3], 2)
True
>>> has_consecutive_duplicates_of([2, 0, 2], 2)
False
"""
if not seq:
return False
for i in range(len(seq) - 1):
if seq[i] == item and seq[i + 1] == item:
return True
return False
def has_duplicates_of(seq, item):
"""Returns whether seq contains duplicates of the given item.
Args:
seq: The sequence.
item: Which item to test for duplicates.
>>> has_duplicates_of([], 5)
False
>>> has_duplicates_of([1, 2, 3], 5)
False
>>> has_duplicates_of([1, 2, 3], 2)
False
>>> has_duplicates_of([1, 2, 2, 3], 2)
True
"""
return len(list(x for x in seq if x == item)) > 1
def count_items(seq, item):
"""Returns the number of times the given item appears in the sequence.
Args:
seq: The sequence.
item: Which item to return the count of.
>>> count_items([], 5)
0
>>> count_items([1, 2, 3], 5)
0
>>> count_items([1, 2, 3], 2)
1
>>> count_items([1, 2, 3, 2], 2)
2
"""
return len(list(x for x in seq if x == item))
def slice(seq, count=2):
"""Splits a sequence into a list of count sublists.
If count > len(seq) the returned list will still contain count sublists, but
len(seq) - count of the sublists will be empty.
From http://www.garyrobinson.net/2008/04/splitting-a-pyt.html
Args:
seq: The sequence.
count: The number of sublists.
>>> slice([], 2)
[[], []]
>>> slice([1, 2], 1)
[[1, 2]]
>>> slice([1, 2], 2)
[[1], [2]]
>>> slice(range(6), 2)
[[0, 1, 2], [3, 4, 5]]
>>> slice(range(5), 2)
[[0, 1, 2], [3, 4]]
>>> slice([1, 2], 3)
[[1], [2], []]
"""
start = 0
output = []
for i in xrange(count):
stop = start + len(seq[i::count])
output.append(seq[start:stop])
start = stop
return output
if __name__ == "__main__":
import doctest
doctest.testmod()
|
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Comments
A shorthand for
in Python is:
If your sequences are really long, there's a trick to avoid using up lots of memory when counting items. Instead of
it's probably better to use
#
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