Question

What is the pythonic way to test if there is a tuple starting with another tuple in collection? actually, I am really after the index of match, but I can probably figure out from test example

for example:

c = ((0,1),(2,3))
# (0,) should match first element, (3,)should match no element

I should add my python is 2.4 and/or 2.5

thanks

Answer 1

Edit:
Thanks to the OP for the addition explanation of the problem.
S.Mark’s nested list comprehensions are pretty wicked; check ’em out.

I might opt to use an auxiliary function:

def tup_cmp(mytup, mytups):
    return any(x for x in mytups if mytup == x[:len(mytup)])

>>> c = ((0, 1, 2, 3), (2, 3, 4, 5))
>>> tup_cmp((0,2),c)
False
>>> tup_cmp((0,1),c)
True
>>> tup_cmp((0,1,2,3),c)
True
>>> tup_cmp((0,1,2),c)
True
>>> tup_cmp((2,3,),c)
True
>>> tup_cmp((2,4,),c)
False

---

Original answer:
Does using a list-comprehension work for you?:

c = ((0,1),(2,3))

[i for i in c if i[0] == 0]
# result: [(0, 1)]

[i for i in c if i[0] == 3]
# result: []

List comps were introduced in 2.0.