问题
Python's succint syntax through its batteries allows verbose code line to be expressed in readable one liners. Consider the following examples
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for a in range(3): |
for b in range(3): |
for c in range(3): |
print (a,b,c), |
- - - - - - - - - - - - - - - - - -|
for e in product(range(3), repeat=3): |
print e, |
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for a in range(3): |
for b in range(a , 3): |
for c in range(b , 3): |
print (a,b,c), |
- - - - - - - - - - - - - - - - - -|
for e in combinations_with_replacement(range(3), 3):|
print e, |
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for a in range(3): |
for b in range(a + 1, 3): |
for c in range(b + 1, 3): |
print (a,b,c), |
- - - - - - - - - - - - - - - - - -|
for e in combinations(range(3), 3): |
print e, |
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for a in range(3): |
for b in range(3): |
for c in range(3): |
if len(set([a,b,c])) == 3: |
print (a,b,c), |
- - - - - - - - - - - - - - - - - -|
for e in permutations(range(3)): |
print e, |
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Of Late I ended up with a deep nested dependent Loop I was trying to express succinctly but failed
The structure of the loop would be as follows
for a in A():
for b in B(a):
for c in C(b):
foo(a,b,c)
Can such structure be expressed in an equivalent itertools notation?
回答1:
There's no exact itertools solution, but a simple combination of itertools functions will suffice:
def chain_imap_accumulate(seq, f):
def acc_f(x):
for n in f(x[-1]):
yield x + (n,)
return chain.from_iterable(imap(acc_f, seq))
def accumulative_product(*generators):
head, tail = generators[0], generators[1:]
head = imap(tuple, head())
return reduce(chain_imap_accumulate, tail, head)
A quick test. Definitions:
from itertools import chain, imap, izip
chain_ = chain.from_iterable
def A():
yield 'A'
yield 'B'
def B(x):
yield int(x, 16)
yield int(x, 16) + 1
def C(x):
yield str(x) + 'Z'
yield str(x) + 'Y'
And the result:
>>> list(accumulative_product(A, B, C))
[('A', 10, '10Z'), ('A', 10, '10Y'),
('A', 11, '11Z'), ('A', 11, '11Y'),
('B', 11, '11Z'), ('B', 11, '11Y'),
('B', 12, '12Z'), ('B', 12, '12Y')]
Almost all the complexity comes from the accumulation of inputs, as a quick "derivation" of the above code shows. The final (c) values can be generated using just a couple of nested itertools constructs:
>>> list(chain_(imap(C, chain_(imap(B, (A()))))))
['10Z', '10Y', '11Z', '11Y', '11Z', '11Y', '12Z', '12Y']
This can be generalized with reduce. To work with reduce, chain_imap can't use the standard imap argument order. It has to be swapped:
def chain_imap(seq, f):
return chain.from_iterable(imap(f, seq))
This gives the same results:
>>> list(reduce(chain_imap, [B, C], A()))
['10Z', '10Y', '11Z', '11Y', '11Z', '11Y', '12Z', '12Y']
The final task is accumulating the initial values, so that you have access to a, b, and c. This takes a bit of thought to get right, but the implementation is fairly simple -- we just have to convert f into a function that ignores all input values but the last, and appends new values to the full input:
def chain_imap_accumulate(seq, f):
def acc_f(x):
for n in f(x[-1]):
yield x + (n,)
return chain.from_iterable(imap(acc_f, seq))
This requires that the first inputs be wrapped in tuples, so we map A with tuple:
>>> list(reduce(chain_imap_accumulate, [B, C], imap(tuple, A())))
[('A', 10, '10Z'), ('A', 10, '10Y'),
('A', 11, '11Z'), ('A', 11, '11Y'),
('B', 11, '11Z'), ('B', 11, '11Y'),
('B', 12, '12Z'), ('B', 12, '12Y')]
Rewrite the above for clarity, and the code at the top of this answer results.
By the way, chain_imap_accumulate can be rewritten a bit more tersely using a genex. This can be combined with a shorter version of accumulative_product for a very compact definition (if you're interested in that kind of thing). This also happens to eliminate the itertools dependency entirely:
def chain_map_accumulate(seq, f):
return (x + (n,) for x in seq for n in f(x[-1]))
def accumulative_product2(*gens):
return reduce(chain_map_accumulate, gens[1:], (tuple(x) for x in gens[0]()))
回答2:
There is not, but you can make one:
def chainGang(steps, currentVars=None):
thisOne = steps[0]
if currentVars is None:
for item in thisOne():
for gang in chainGang(steps[1:], [item]):
yield gang
elif len(steps) == 1:
for item in thisOne(currentVars[-1]):
yield currentVars + [item]
else:
for item in thisOne(currentVars[-1]):
for gang in chainGang(steps[1:], currentVars + [item]):
yield gang
And then:
>>> outer = lambda: ["A", "B", "C", "D"]
>>> middle = lambda letter: [letter, letter*2, letter*3]
>>> inner = lambda s: range(len(s)+1)
>>> for a in chainGang([outer, middle, inner]):
... print a
[u'A', u'A', 0]
[u'A', u'A', 1]
[u'A', u'AA', 0]
[u'A', u'AA', 1]
[u'A', u'AA', 2]
[u'A', u'AAA', 0]
[u'A', u'AAA', 1]
[u'A', u'AAA', 2]
[u'A', u'AAA', 3]
[u'B', u'B', 0]
[u'B', u'B', 1]
[u'B', u'BB', 0]
[u'B', u'BB', 1]
[u'B', u'BB', 2]
[u'B', u'BBB', 0]
[u'B', u'BBB', 1]
[u'B', u'BBB', 2]
[u'B', u'BBB', 3]
[u'C', u'C', 0]
[u'C', u'C', 1]
[u'C', u'CC', 0]
[u'C', u'CC', 1]
[u'C', u'CC', 2]
[u'C', u'CCC', 0]
[u'C', u'CCC', 1]
[u'C', u'CCC', 2]
[u'C', u'CCC', 3]
[u'D', u'D', 0]
[u'D', u'D', 1]
[u'D', u'DD', 0]
[u'D', u'DD', 1]
[u'D', u'DD', 2]
[u'D', u'DDD', 0]
[u'D', u'DDD', 1]
[u'D', u'DDD', 2]
[u'D', u'DDD', 3]
来源：https://stackoverflow.com/questions/15037175/equivalent-nested-loop-structure-with-itertools