问题
Can someone explain the structure of reduce() in the following example:
def f2(list):
return reduce(lambda string, item: string + chr(item), list, "")
I know that f2 converts a list of int's into a string, but my problem is understanding reduce in this context. I know the basic structure of reduce is reduce(function, sequence[, initial]) but this is somehow confusing to me. Can someone explain reduce(lambda string, item: string + chr(item), list, "") and give me some similar examples ? Thanks in advance.
回答1:
The code applies chr() to every element of the list, and concatenates the results into a single string.
The reduce() call is equivalent to the following:
return "" + chr(list[0]) + chr(list[1]) + ... + chr(list[list.length - 1])
The "" is the third argument to reduce(). The lambda function in
return reduce(lambda string, item: string + chr(item), list, "")
is called for every item in the list. It simply appends chr(item) to the result of the previous iteration.
For more examples of using reduce(), see Useful code which uses reduce() in python
回答2:
return reduce(lambda string, item: string + chr(item), list, "")
roughly translates to
string = ""
for item in list:
string = string + chr(item)
return string
回答3:
Reduce does something usually called a fold. E.g., if you have a list ls = [a,b,c,d] and a binary operation def plus(x,y): x + y, then reduce(plus, ls) gets folded to
plus(plus(plus(a, b), c), d)
which equals
(((a+b)+c)+d)
Your f2 is doing something similar, namely appending strings (after converting them from integers): (I really hope those parens match...)
(((("" + chr(a)) + chr(b)) + chr(c)) + chr(d))
with a supplied initial value of "" (which is needed when a folding operation has two different input types)
@ python experts: I'm not sure if reduce is a left fold, it seemed more naturally to me. Please tell me if I'm wrong.
来源：https://stackoverflow.com/questions/10429079/understanding-complex-code-with-reduce-python