CODEWITHSUNDEEP
pythonpython-2.7
skyork
skyork

Reputation: 7381

Create a list out of a tuple of tuples

I have a tuple of tuples like this: t = ((4, 3), (2, 9), (7, 2), ...), where the first element in each nested tuple (i.e. t[i][0]) can range from 1 to 11 without repetition, but not necessarily every integer between 1 and 11 will be present.

I want to create a list (or tuple) r based on t, in the following way:

The resulting list r has a length of 11 exactly. For each index j in r, if j + 1 === t[i][0] for any i, then r[j] = t[i][1], otherwise r[j] = 0.

This can be done by initializing r to [0] * 11 first, and then loop through t to assign t[i][1] to r[t[i][0] - 1]:

t = ((4, 3), (2, 9), (7, 2), (10, 1))
r = [0] * 11
for item in t:
    r[item[0] - 1] = item[1]

r = [0, 9, 0, 3, 0, 0, 2, 0, 0, 1, 0]

But is there any more efficient way (a functional way, maybe)?

Upvotes: 4

Views: 201

Answers (4)

lucasg
lucasg

Reputation: 11002

Just use a mapping (by dictionary)

d = { v[0]:v[1] for i,v in enumerate(t) }
result = [ d[j+1] if j+1 in d else 0  for j in t ]

Upvotes: 1

Ashwini Chaudhary
Ashwini Chaudhary

Reputation: 250881

You can use dict and a list comprehension with conditional expression:

>>> dic = dict(t)
>>> [dic[i] if i in dic else 0 for i in range(1, 12)]
[0, 9, 0, 3, 0, 0, 2, 0, 0, 1, 0]

Upvotes: 1

Jon Clements
Jon Clements

Reputation: 142106

How about:

>>> t
((4, 3), (2, 9), (7, 2), (10, 1))
>>> d = dict(t)
>>> [d.get(el, 0) for el in xrange(1, 12)]
[0, 9, 0, 3, 0, 0, 2, 0, 0, 1, 0]

Upvotes: 6

iruvar
iruvar

Reputation: 23374

I would create a dictionary from t and populate r using lookups

t = ((4, 3), (2, 9), (7, 2))
d_t = dict(t)
r = [0]*11
r = [d_t[i+1] if i + 1 in d_t else r[i] for i, x in enumerate(r)]
print r
[0, 9, 0, 3, 0, 0, 2, 0, 0, 0, 0]

Upvotes: 1

Related Questions