Python list comprehension [duplicate]

This question already has answers here: 开发者_运维技巧 Closed 12 years ago.

Possible Duplicate:

List comprehension for running total

I'm trying to write a concise list comprehension statement to create a cdf: For example:

print f([0.2, 0.3,0.1,0.4])
[0.2,0.5,0.6,1.0] 

A standard procedure would look like this (I want to write a list comprehension for the function f()):

def f(probabilities) :

    sum = 0
    returnList = []
    for count in probabilities:
        sum +=count
        returnList = returnList + [sum]
    return returnList

Edit: I found a function numpy.cumsum(). I'll check if it uses list comprehensions.

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That operation is so common that many languages (mainly functional ones, but not only) provide abstractions for it, usually with the name scanl (it's like a reduce with intermediate results). Let's call it ireduce ("iterative reduce"):

def ireduce(f, state, it):
    for x in it:
        state = f(state, x)
        yield state

And now use it:

import operator

def f(probabilities):
    return ireduce(operator.add, 0, probabilities)

print(list(f([0.2, 0.3,0.1,0.4])))
# [0.2, 0.5, 0.6, 1.0]

---

[sum(probabilities[:i+1]) for i in range(len(probabilities))]

But don't do that because it's O(n^2). Python list comprehensions weren't designed for this. Use the procedural code that you already wrote.

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It's not really pretty, and it's not using list comprehensions, but you can do this with the reduce() function, where the accumulated value is a tuple holding the current sum and the result list:

a = [0.2, 0.3, 0.1, 0.4]
reduce((lambda result, val: (result[0] + val, result[1] + [result[0] + val])), a, (0, []))[1]

Python's lack of support for multi-line lambda's makes this kind of ugly. Using a separate function would be better:

    a = [0.2, 0.3, 0.1, 0.4]   
    def accumulate(result, val):
        return (result[0] + val, result[1] + [result[0] + val])

    reduce(accumulate, a, (0, []))[1]