The power of Python's yield

Permutation is the rearrangement of objects or symbols into distinguishable sequences. Each unique ordering is called a permutation. For example, with the numerals one to six, each possible ordering consists of a complete list of the numerals, without repetitions. There are 720 total permutations of these numerals, one of which is: "4, 5, 6, 1, 2, 3".

How do you go about calculating all possible permutations of a set? Well, if you know a priori the size of the input set, you can just write as many loops as is the set's cardinality. If, for example, we only have to deal with sets of 4 elements, ...

for a in (1,2,3,4): for b in (1,2,3,4): if b == a: continue # No repetitions allowed for c in (1,2,3,4): if c in [a,b]: continue # No repetitions allowed for d in (1,2,3,4): if d in [a,b,c]: continue # No repetitions allowed print values[a], values[b], values[c], values[d]

That's all fine and simple, but what if you don't know in
advance how large the set is?

You have to perform the same kind of loops, only you don't
know how deep you'll have to go.

"Deep"?

Whenever "deep" comes up, recursion is knocking on the door...

def permute(inputData, outputSoFar=[]): outputSizeToReach = len(inputData) for elem in inputData: if elem not in outputSoFar: outputSoFar.append(elem) if len(outputSoFar) == outputSizeToReach: print outputSoFar else: permute(inputData, outputSoFar) # --- Recursion outputSoFar.pop() permute([1, 2, 3])

$ python simple.py [1, 2, 3] [1, 3, 2] [2, 1, 3] [2, 3, 1] [3, 1, 2] [3, 2, 1]

Clean, simple code. But... It just prints the results. What if we want to take some action on each permutation?

We could of course pass a lambda or a function (for execution on each permutation) as an extra argument to permute... But there is a better, a much better way:

def permute(inputData, outputSoFar=[]): outputSizeToReach = len(inputData) for elem in inputData: if elem not in outputSoFar: outputSoFar.append(elem) if len(outputSoFar) == outputSizeToReach: yield outputSoFar else: for v in permute(inputData, outputSoFar): yield v outputSoFar.pop()

for i in permute([1,2,3]): print i # or whatever else you want to do with it

In case you haven't met `yield` before: this requires no
extra runtime memory! The iteration process uses the magic
behind yield, which "freezes" the state (call stack,
"instruction pointer", etc)
when it returns the result, and continues from the exact
place it last was, when the iterator protocol asks
for the next value.

Note that the code above doesn't use any kind of library... It
simply uses a language feature to implement what we need.
Now try implementing the same permutation interface in plain C++,
without using any library (STL permutations or otherwise) ...
See why I love `yield`? :‑)

*P.S. Notice that where the original permute
function called itself (the recursive call) a slight
modification had to be done: by invoking yield, the permute
function is now a generator, so we can't just simply
call it again - if we did, we would lose the returned
sequence, and thus, the returned permutations! We must
instead loop over the returned results with a simple for v in
permute(...): yield v. This way, the results are properly
propagated to the "parent" call.*

** Update:** For people developing in C/C++, here is
an excellent article by Simon Tatham
on how he implemented coroutines (i.e. yield-like behaviour) in C.

** Update, some years later:** It's also important to know
that there's a difference between how Python implements yield
and how other languages do.

Index CV | Updated: Sat Oct 8 12:33:59 2022 |

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