def partition_all(n, seq):
""" Partition all elements of sequence into tuples of length at most n
The final tuple may be shorter to accommodate extra elements.
>>> list(partition_all(2, [1, 2, 3, 4]))
[(1, 2), (3, 4)]
>>> list(partition_all(2, [1, 2, 3, 4, 5]))
[(1, 2), (3, 4), (5,)]
See Also:
partition
"""
args = [iter(seq)] * n
it = zip_longest(*args, fillvalue=no_pad)
try:
prev = next(it)
except StopIteration:
return
for item in it:
yield prev
prev = item
if prev[-1] is no_pad:
yield prev[:prev.index(no_pad)]
else:
yield prev
def partition_all(n, seq):
""" Partition all elements of sequence into tuples of length at most n
The final tuple may be shorter to accommodate extra elements.
>>> list(partition_all(2, [1, 2, 3, 4]))
[(1, 2), (3, 4)]
>>> list(partition_all(2, [1, 2, 3, 4, 5]))
[(1, 2), (3, 4), (5,)]
See Also:
partition
"""
args = [iter(seq)] * n
it = zip_longest(*args, fillvalue=no_pad)
try:
prev = next(it)
except StopIteration:
return
for item in it:
yield prev
prev = item
if prev[-1] is no_pad:
yield prev[:prev.index(no_pad)]
else:
yield prev
def partition_all(n, seq):
""" Partition all elements of sequence into tuples of length at most n
The final tuple may be shorter to accommodate extra elements.
>>> list(partition_all(2, [1, 2, 3, 4]))
[(1, 2), (3, 4)]
>>> list(partition_all(2, [1, 2, 3, 4, 5]))
[(1, 2), (3, 4), (5,)]
See Also:
partition
"""
args = [iter(seq)] * n
it = zip_longest(*args, fillvalue=no_pad)
try:
prev = next(it)
except StopIteration:
return
for item in it:
yield prev
prev = item
if prev[-1] is no_pad:
try:
# If seq defines __len__, then
# we can quickly calculate where no_pad starts
yield prev[:len(seq) % n]
except TypeError:
# Get first index of no_pad without using .index()
# https://github.com/pytoolz/toolz/issues/387
# Binary search from CPython's bisect module,
# modified for identity testing.
lo, hi = 0, n
while lo < hi:
mid = (lo + hi) // 2
if prev[mid] is no_pad:
hi = mid
else:
lo = mid + 1
yield prev[:lo]
else:
yield prev
def diff(*seqs, **kwargs):
""" Return those items that differ between sequences
>>> list(diff([1, 2, 3], [1, 2, 10, 100]))
[(3, 10)]
Shorter sequences may be padded with a ``default`` value:
>>> list(diff([1, 2, 3], [1, 2, 10, 100], default=None))
[(3, 10), (None, 100)]
A ``key`` function may also be applied to each item to use during
comparisons:
>>> list(diff(['apples', 'bananas'], ['Apples', 'Oranges'], key=str.lower))
[('bananas', 'Oranges')]
"""
N = len(seqs)
if N == 1 and isinstance(seqs[0], list):
seqs = seqs[0]
N = len(seqs)
if N < 2:
raise TypeError('Too few sequences given (min 2 required)')
default = kwargs.get('default', no_default)
if default == no_default:
iters = zip(*seqs)
else:
iters = zip_longest(*seqs, fillvalue=default)
key = kwargs.get('key', None)
if key is None:
for items in iters:
if items.count(items[0]) != N:
yield items
else:
for items in iters:
vals = tuple(map(key, items))
if vals.count(vals[0]) != N:
yield items
def diff(*seqs, **kwargs):
""" Return those items that differ between sequences
>>> list(diff([1, 2, 3], [1, 2, 10, 100]))
[(3, 10)]
Shorter sequences may be padded with a ``default`` value:
>>> list(diff([1, 2, 3], [1, 2, 10, 100], default=None))
[(3, 10), (None, 100)]
A ``key`` function may also be applied to each item to use during
comparisons:
>>> list(diff(['apples', 'bananas'], ['Apples', 'Oranges'], key=str.lower))
[('bananas', 'Oranges')]
"""
N = len(seqs)
if N == 1 and isinstance(seqs[0], list):
seqs = seqs[0]
N = len(seqs)
if N < 2:
raise TypeError('Too few sequences given (min 2 required)')
default = kwargs.get('default', no_default)
if default == no_default:
iters = zip(*seqs)
else:
iters = zip_longest(*seqs, fillvalue=default)
key = kwargs.get('key', None)
if key is None:
for items in iters:
if items.count(items[0]) != N:
yield items
else:
for items in iters:
vals = tuple(map(key, items))
if vals.count(vals[0]) != N:
yield items
def partition(n, seq, pad=no_pad):
""" Partition sequence into tuples of length n
>>> list(partition(2, [1, 2, 3, 4]))
[(1, 2), (3, 4)]
If the length of ``seq`` is not evenly divisible by ``n``, the final tuple
is dropped if ``pad`` is not specified, or filled to length ``n`` by pad:
>>> list(partition(2, [1, 2, 3, 4, 5]))
[(1, 2), (3, 4)]
>>> list(partition(2, [1, 2, 3, 4, 5], pad=None))
[(1, 2), (3, 4), (5, None)]
See Also:
partition_all
"""
args = [iter(seq)] * n
if pad is no_pad:
return zip(*args)
else:
return zip_longest(*args, fillvalue=pad)
def partition(n, seq, pad=no_pad):
""" Partition sequence into tuples of length n
>>> list(partition(2, [1, 2, 3, 4]))
[(1, 2), (3, 4)]
If the length of ``seq`` is not evenly divisible by ``n``, the final tuple
is dropped if ``pad`` is not specified, or filled to length ``n`` by pad:
>>> list(partition(2, [1, 2, 3, 4, 5]))
[(1, 2), (3, 4)]
>>> list(partition(2, [1, 2, 3, 4, 5], pad=None))
[(1, 2), (3, 4), (5, None)]
See Also:
partition_all
"""
args = [iter(seq)] * n
if pad is no_pad:
return zip(*args)
else:
return zip_longest(*args, fillvalue=pad)