Question

是否有像F# sSeq.scan()这样的功能?

我想做的是<代码>和()或cumproduct()类无照物品。

Answer 1

Ignacio的解决方案几乎是我所认为的,但需要一种类型的操作者(a->a->a)并且不会产生第一种要素。

def scan(f, state, it):
  for x in it:
    state = f(state, x)
    yield state
# test
>>> snoc = lambda xs,x: xs+[x]
>>> list(scan(snoc, [],  abcd ))
[[ a ], [ a ,  b ], [ a ,  b ,  c ], [ a ,  b ,  c ,  d ]]
>>> list(scan(operator.add, 0, [1,2,3]))
[1,3,6]

具体来说,<代码>Seq.scan的类型。

( State ->  T ->  State) ->  State -> seq< T> -> seq< State>

落成型号为scan。

( State ->  State ->  State) -> seq< State> -> seq< State>

具体来说,这源自于斯图尔特(reduce<>/code(>)的行文。

Answer 2

Nope。

def scan(op, seq):
  it = iter(seq)
  result = next(it)
  for val in it:
    result = op(result, val)
    yield result

Answer 3

如果仅涉及cum/积业务,那么你就应当看一看一看一 efficient的超高效率cumsum 和 cumprod 。

Answer 4

汇总职能将使用<条码>代谢<>。页: 1

见 rel=“nofollow noreferer” http://docs.python.org/library/Functions.html。更多信息

Answer 5

您可使用累积功能,其功能与扫描等同:
。

from itertools import accumulate
nums=[1,2,3,4,5]
nums_added=list(accumulate(nums,lambda x,y:x+y))
print(nums_added)

#prints
# [1,3,6,10,15]

You can also use rx (reactive extension) to use scan operator. It s a external library. so you need to install it before using it. with rx you can do something like this:

from rx import  from_, operators as op
nums=[1,2,3,4,5]
source=from_(nums).pipe(op.scan(lambda x,y:x+y))
source.subscribe(lambda num:print(num)

#prints
#1
#3
#6
#10
#15

rx can do much more than this it allows you to do reactive programming. To learn more visit this: http://reactivex.io

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