Sort generated numbers using another python generator

generator, python, sorting

Solution

You are replacing your `(value, index)` tuples with just the value:

self.values[index] = self.generators[index].next()

You need to replace that with a new tuple:

self.values[index] = (self.generators[index].next(), index)

otherwise the iterable assignment fails; you cannot assign one `int` to two variables.

Your generator is missing a loop and handling of empty generators:

def generate(self):
    while any(self.values):
        r, index = min(v for v in self.values if v)
        try:
            self.values[index] = (self.generators[index].next(), index)
        except StopIteration:
            self.values[index] = None
        yield r

This sets elements of your `self.values` list to `None` to indicate the iterable has been exhausted. This is not the most efficient way to handle this edge case; in a version I wrote before I used a dictionary to track active iterables and simply deleted from that to keep indices (keys) stable.

Note that you can replace your `t()` function with the built-in `iter()` function.

Demo:

>>> class GeneratorSort():
...     def __init__(self, *args):
...         self.values = [(arg.next(), i) for i, arg in enumerate(args)]
...         self.generators = args
...     def generate(self):
...         while any(self.values):
...             r, index = min(v for v in self.values if v)
...             try:
...                 self.values[index] = (self.generators[index].next(), index)
...             except StopIteration:
...                 self.values[index] = None
...             yield r
... 
>>> l1 = [2, 5, 6, 8]
>>> l2 = [1, 4, 5, 7]
>>> l3 = [0, 3, 9, 10]
>>> a = GeneratorSort(iter(l1), iter(l2), iter(l3))
>>> list(a.generate())
[0, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10]

The standard library does it more efficiently still with the `heapq.merge()` function; it uses a heap to keep the iterables sorted by lowest value in a very efficient manner; `min()` needs to loop through all K iterables, while using a heap only takes log-K steps to keep the heap invariant intact.

>>> import heapq
>>> list(heapq.merge(l1, l2, l3))
[0, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10]

You can study the source code, which has been highly tuned for maximum performance.

Problem

Original source

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