Python List Comprehension. Generating a List by repeatedly calling a function till it returns a predefined value

Question

I would like to generate a list by repeatedly calling a function till the function return a predefined value.

myList = [i,r = foo() while i==0]

myList = [r1, r2, r3, r4...] as long as returned i value is == 0

Is there a way to do this using list comprehension?

Is a list comprehension truly a requirement? Seems like this would be very easily written and read as a generator, for example. — hexparrot, Oct 30 '13 at 15:39
Where are you getting the `r` values from? Does your function return two values? — arshajii, Oct 30 '13 at 15:39
Possible duplicate http://stackoverflow.com/questions/9572833/break-list-comprehension — Benjamin Toueg, Oct 30 '13 at 15:39
Generator is also fine. The function foo() returns two values. i,r. — Pavan, Oct 30 '13 at 15:41
foo is not a generator function. It returns two different values when ever we call it i.e. i,r. — Pavan, Oct 30 '13 at 15:48

Benjamin Toueg · Accepted Answer · 2013-10-31T08:25:47.840

1

def stop_iteration():
    raise StopIteration

If foo() is a generator:

myList = list((i,r) if i != 0 else stop_iteration() for (i,r) in foo())

else:

def foo_generator():
    i,r = foo()
    while i != 0:
      yield i,r
      i,r = foo()

myList = list(foo_generator())

edited Oct 31 '13 at 08:25

answered Oct 30 '13 at 15:44

Benjamin Toueg

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score 0 · Answer 2 · answered Oct 30 '13 at 15:51

I don't think there is a one-liner for that. However, by defining a simple generator which would wrap any given foo-like function, you can do it.

TEST = [(0, 0), (1, 0), (2, 1), (3, 0)]


def foo():
    return TEST.pop(0)


def foo_wrapper(foo_like):
    r, i = foo_like()
    while i == 0:
        yield r
        r, i = foo_like()


print list(foo_wrapper(foo))

score 0 · Answer 3 · answered Oct 30 '13 at 15:57

No, you cannot do that. The only keywords that list comprehensions support are for, if, and else (and lambda of course).

So, you have to do a multi-line solution. However there is no need to define a wrapper function. This will work fine:

mylist = []
while True:   
    i, r = foo()
    if i: break
    mylist.append(r)

In the end, mylist will have everything you want.

score 0 · Answer 4 · answered Oct 30 '13 at 16:08

This is not recommended, but...

>>> def notyet(predicate):
        if predicate:
            raise StopIteration
        return True

>>> # note that below we're wrapping a generator expression in a 
>>> # call to `list` so that when `StopIteration` is raised your
>>> # program won't terminate.
>>> xs = list((i, r) for (i, r) in foo() if notyet(i!=0))
>>> # if we try this instead, your program will hit the buffers
>>> # when the predicate is satisfied.
>>> xs = [(i, r) for (i, r) in foo() if notyet(i!=0)]
Traceback (most recent call last):
  File "<pyshell#5>", line 1, in <module>
    xs = [(i, r) for (i, r) in foo() if notyet(i!=0)]
  File "<pyshell#3>", line 3, in notyet
raise StopIteration
StopIteration

It's much better to use itertools.takewhile in this situation.

>>> from itertools import takewhile
>>> xs = list(takewhile(lambda (i, r): i!=0, foo()))

Python List Comprehension. Generating a List by repeatedly calling a function till it returns a predefined value

4 Answers4