Pattern matching of lists in Python Pattern matching of lists in Python python python

Pattern matching of lists in Python


So far as I know there's no way to make it a one-liner in current Python without introducing another function, e.g.:

split_list = lambda lst: (lst[0], lst[1:])head, rest = split_list(my_func())

However, in Python 3.0 the specialized syntax used for variadic argument signatures and argument unpacking will become available for this type of general sequence unpacking as well, so in 3.0 you'll be able to write:

head, *rest = my_func()

See PEP 3132 for details.


First of all, please note that the "pattern matching" of functional languages and the assignment to tuples you mention are not really that similar. In functional languages the patterns are used to give partial definitions of a function. So f (x : s) = e does not mean take the head and tail of the argument of f and return e using them, but it means that if the argument of f is of the form x : s (for some x and s), then f (x : s) is equal to e.

The assignment of python is more like a multiple assignment (I suspect that was its original intention). So you write, for example, x, y = y, x to swap the values in x and y without needing a temporary variable (as you would with a simple assignment statement). This has little to do with pattern matching as it is basically a shorthand for the "simultaneous" execution of x = y and y = x. Although python allows arbitrary sequences instead of comma-separated lists, I would not suggest calling this pattern matching. With pattern matching you check whether or not something matches a pattern; in the python assignment you should ensure that the sequences on both sides are the same.

To do what you seem to want you would usually (also in functional languages) use either a auxiliary function (as mentioned by others) or something similar to let or where constructs (which you can regard as using anonymous functions). For example:

(head, tail) = (x[0], x[1:]) where x = my_func()

Or, in actual python:

(head, tail) = (lambda x: (x[0], x[1:]))(my_func())

Note that this is essentially the same as the solutions given by others with an auxiliary function except that this is the one-liner you wanted. It is, however, not necessarily better than a separate function.

(Sorry if my answer is a bit over the top. I just think it's important to make the distinction clear.)


That's a very much a 'pure functional' approach and as such is a sensible idiom in Haskell but it's probably not so appropriate to Python. Python only has a very limited concept of patterns in this way - and I suspect you might need a somewhat more rigid type system to implement that sort of construct (erlang buffs invited to disagree here).

What you have is probably as close as you would get to that idiom, but you are probably better off using a list comprehension or imperative approach rather than recursively calling a function with the tail of the list.

As has been stated on a few occasions before, Python is not actually a functional language. It just borrows ideas from the FP world. It is not inherently Tail Recursive in the way you would expect to see embedded in the architecture of a functional language, so you would have some difficulty doing this sort of recursive operation on a large data set without using a lot of stack space.