I have two array like this:
[('a', 'beta'), ('b', 'alpha'), ('c', 'beta'), .. ]
[('b', 37), ('c', 22), ('j', 93), .. ]
I want to produce something like:
[('b', 'alpha', 37), ('c', 'beta', 22), .. ]
Is there an easy way to do this?
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1have you tried anything?depperm– depperm2017-05-09 12:55:13 +00:00Commented May 9, 2017 at 12:55
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@depperm I considered a for loop to check if matching and push to a new array but I thought there might be some built in functions that could make it easier.Philip Kirkbride– Philip Kirkbride2017-05-09 12:56:25 +00:00Commented May 9, 2017 at 12:56
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2Check this thread: stackoverflow.com/questions/7776907/…Cleared– Cleared2017-05-09 12:57:57 +00:00Commented May 9, 2017 at 12:57
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stackoverflow.com/questions/17682721/…Serge– Serge2017-05-09 13:13:24 +00:00Commented May 9, 2017 at 13:13
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2 Answers
I would suggest a hash discriminator join like method:
l = [('a', 'beta'), ('b', 'alpha'), ('c', 'beta')]
r = [('b', 37), ('c', 22), ('j', 93)]
d = {}
for t in l:
d.setdefault(t[0], ([],[]))[0].append(t[1:])
for t in r:
d.setdefault(t[0], ([],[]))[1].append(t[1:])
from itertools import product
ans = [ (k,) + l + r for k,v in d.items() for l,r in product(*v)]
results in:
[('c', 'beta', 22), ('b', 'alpha', 37)]
This has lower complexity closer to O(n+m) than O(nm) because it avoids computing the product(l,r) and then filtering as the naive method would.
Mostly from: Fritz Henglein's Relational algebra with discriminative joins and lazy products
It can also be written as:
def accumulate(it):
d = {}
for e in it:
d.setdefault(e[0], []).append(e[1:])
return d
l = accumulate([('a', 'beta'), ('b', 'alpha'), ('c', 'beta')])
r = accumulate([('b', 37), ('c', 22), ('j', 93)])
from itertools import product
ans = [ (k,) + l + r for k in l&r for l,r in product(l[k], r[k])]
This accumulates both lists separately (turns [(a,b,...)] into {a:[(b,...)]}) and then computes the intersection between their sets of keys. This looks cleaner. if l&r is not supported between dictionaries replace it with set(l)&set(r).
Comments
There is no built in method. Adding package like numpy will give extra functionalities, I assume.
But if you want to solve it without using any extra packages, you can use a one liner like this:
ar1 = [('a', 'beta'), ('b', 'alpha'), ('c', 'beta')]
ar2 = [('b', 37), ('c', 22), ('j', 93)]
final_ar = [tuple(list(i)+[j[1]]) for i in ar1 for j in ar2 if i[0]==j[0]]
print(final_ar)
Output:
[('b', 'alpha', 37), ('c', 'beta', 22)]
