4
I have the following vectors:
a = [('FI', 'SE'), ('SE', 'DK'), ('DK', 'DE'), ('DE', 'PL'), ('PL', 'BY')]
b = [['NL', 'DK', 0], ['NL', 'DE', 0], ['NL', 'BE', 0], ['FI', 'SE', 0.054]]
I need to go through the vector to and locate it on the vector b.
My exit has to return:
c = ['FI', 'SE', 0.054]
How can I do that?
Grateful for the help.
2
If, instead of a list of lists, you can upload b as a dictionary (or transform b in a dictionary bd, as I did in the example below) the code can become more intuitive, just one bd.get to locate the required value. Ex.:
a = [('FI', 'SE'), ('SE', 'DK'), ('DK', 'DE'), ('DE', 'PL'), ('PL', 'BY')]
b = [['NL', 'DK', 0], ['NL', 'DE', 0], ['NL', 'BE', 0], ['FI', 'SE', 0.054]]
bd = dict(((x, y), z) for x, y, z in b)
print(bd)
# saída: {('FI', 'SE'): 0.054, ('NL', 'BE'): 0, ('NL', 'DE'): 0, ('NL', 'DK'): 0}
c = [(i, bd.get(i)) for i in a if i in bd]
print(c)
# saída: [(('FI', 'SE'), 0.054)]
@Miguel, actually this code is much faster than what you proposed, even adding the conversion time of b to a dictionary, because the in in dicts is O(1), while in lists and tuples is O(n).
With small datasets the difference is derisory (milliseconds in favor of the A method, without a dictionary), but with large datasets the difference is orders of magnitude in favor of the dictionary method: https://repl.it/WorthlessAustereRegression
@Miguel It is part of learning, for everyone. : ) I didn’t need to delete the comments, they might be useful for other people who had the same doubt about the approaches.
It was already very messy here
2
You can try it like this:
a = [('FI', 'SE'), ('SE', 'DK'), ('DK', 'DE'), ('DE', 'PL'), ('PL', 'BY')]
b = [['NL', 'DK', 0], ['NL', 'DE', 0], ['NL', 'BE', 0], ['FI', 'SE', 0.054]]
search1, search2 = ('FI', 'SE')
for val1, val2, val3 in b:
if((val1, val2) == (search1, search2)):
print(val1, val2, val3) # FI SE 0.054
break
else: # caso nao haja break e porque nao foi encontrado
print('não foi encontrado')
If you want to locate all elements of a in b, can:
a = [('FI', 'SE'), ('SE', 'DK'), ('DK', 'DE'), ('DE', 'PL'), ('PL', 'BY')]
b = [['NL', 'DK', 0], ['NL', 'DE', 0], ['NL', 'BE', 0], ['DE', 'PL', 0], ['FI', 'SE', 0.054]]
founds = []
for val1, val2, val3 in b:
if((val1, val2) in a):
founds.append([val1, val2, val3])
print(founds) # [['DE', 'PL', 0], ['FI', 'SE', 0.054]]
Or:
a = [('FI', 'SE'), ('SE', 'DK'), ('DK', 'DE'), ('DE', 'PL'), ('PL', 'BY')]
b = [['NL', 'DK', 0], ['NL', 'DE', 0], ['NL', 'BE', 0], ['DE', 'PL', 0], ['FI', 'SE', 0.054]]
founds = [[v[0], v[1], v[2]] for v in b if (v[0], v[1]) in a]
print(founds) # [['DE', 'PL', 0], ['FI', 'SE', 0.054]]
Thank you @Miguel. Works perfectly!
You’re welcome @Danilo
1
See if that helps you:
a = [('FI', 'SE'), ('SE', 'DK'), ('DK', 'DE'), ('DE', 'PL'), ('PL', 'BY')]
b = [['NL', 'DK', 0], ['NL', 'DE', 0], ['NL', 'BE', 0], ['FI', 'SE', 0.054]]
result = []
for x in a:
for z in b:
if len(list(set(x) - set(z))) == 0:
result.append(z)
print(result) # [['FI', 'SE', 0.054]]
In the example I go through the 2 lists and check the difference between the elements to get the result you expect, and then add to the list result.
NOTE: I’m starting my studies in python, so sorry about anything.
Thank you @Juniornunes your solution also suits me! (+1)
This approach has the problem of considering ('FI', 'SE') == ('SE', 'FI'), due to the use of sets. In addition, the conversion + operation with sets to each item makes the code super slow (2 minutes to compare two lists with 10,000 items each).
Thanks for the @hdiogenes remark!
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Will the two lists always have the same amount of sublists? If a has a has 5 sublists then b also has? Or it may be different?
– Miguel
They can have different sizes @Miguel.
– Danilo
How is the search? Search for
(FI, SE)that’s it?– Miguel
Exactly @Miguel! I look for (FI, SE), but the return will be ['FI', 'SE', 0.054], as in b.
– Danilo