Recently I wrote a function to generate certain sequences with nontrivial constraints. The problem came with a natural recursive solution. Now it happens that, even for relatively small input, the sequences are several thousands, thus I would prefer to use my algorithm as a generator instead of using it to fill a list with all the sequences.
Here is an example. Suppose we want to compute all the permutations of a string with a recursive function. The following naive algorithm takes an extra argument ‘storage’ and appends a permutation to it whenever it finds one:
def getPermutations(string, storage, prefix=''): if len(string) == 1: storage.append(prefix + string) # <----- else: for i in range(len(string)): getPermutations(string[:i]+string[i+1:], storage, prefix+string[i]) storage = [] getPermutations('abcd', storage) for permutation in storage: print permutation (Please don’t care about inefficiency, this is only an example.)
Now I want to turn my function into a generator, i.e. to yield a permutation instead of appending it to the storage list:
def getPermutations(string, prefix=''): if len(string) == 1: yield prefix + string # <----- else: for i in range(len(string)): getPermutations(string[:i]+string[i+1:], prefix+string[i]) for permutation in getPermutations('abcd'): print permutation This code does not work (the function behaves like an empty generator).
Am I missing something? Is there a way to turn the above recursive algorithm into a generator without replacing it with an iterative one?
Or without an accumulator: