将元组元素与列表元素相匹配

Question

I have a list of tuples named eagles like this

eagles= [("NCMS000","NCMS000"),("NCFP000","NCFP000"),("NCMS00D","NCMS00D"),("NCCS000","NCCS000"),("NCCP000","NCCP000"),("NCMN000","NCMN000"),("NCFN000","NCFN000"),("NP000G0","NP000G0"),("NP000G0","NP000G0"),...

and a list named Result like this:

['"', '"', 'Fe', '1']
['Hola', 'hola', 'I', '1']
['como', 'como', 'CS', '0.999289']
['estas', 'este', 'DD0FP0', '0.97043']
['Bien', 'bien', 'NP00000', '1']
['gracias', 'gracia', 'NCFP000', '1']
['y', 'y', 'CC', '0.999962']
['tu', 'tu', 'DP2CSS', '1']
['yo', 'yo', 'PP1CSN00', '1']
['estoy', 'estar', 'VAIP1S0', '1']
['bien', 'bien', 'RG', '0.902728']
['huevo', 'huevo', 'NCMS000', '0.916667']
['calcio', 'calcio', 'NCMS000', '1']
['leche', 'leche', 'NCFS000', '1']
['proteina', 'proteina', 'NCFS000', '1']
['Francisco', 'francisco', 'NP00000', '1']
['1999', '1999', 'Z', '1']
['"', '"', 'Fe', '1']

I need to create a function to compare the 3rd item of Result list with the eagles 1st item in a kind of continuous loop. If they match, I need to return a list of lists w/ the 4 elements, eg:

 r = [['leche', 'leche', 'NCFS000', '1'],['proteina', 'proteina', 'NCFS000', '1'],['Francisco', 'francisco', 'NP00000', '1']]

what I did so far:

def check(lst):
    return [x[2] for x in lst if (x[2] in y[0] for y in eagles)]

IndexError: list index out of range.

I can't even extract the 3rd element from the list and put it on an empty one

e = [x[0] for x in eagles]
r = [item for item in e if item in Result]
rg =[]
for i in Result:
    rg = i[2]

same error

What can I do? Any suggestion is appreciated.

Answer 1

First, of all, it's probably better to convert the eagles list into a dictionary...

>>> eagles = [("NCMS000","NCMS000"), ("NCFP000","NCFP000"), ...]
>>> eagles_dict = dict(eagles)
>>> print eagles_dict
{'NCFP000': 'NCFP000', 'NCMS000': 'NCMS000', ...}

...to make the lookups simpler and more efficient. Then you can use a simple list comprehension, like...

>>> result = [['"', '"', 'Fe', '1'], ['Hola', 'hola', 'I', '1'], ...]
>>> print [item for item in result if item[2] in eagles_dict]
[['leche', 'leche', 'NCFS000', '1'], ...]

Answer 2

There is probably a more efficient algorithm involving sorting but if you're just doing this once or twice:

UPDATED to take into account the fact that your items don't always have 4 elements.

eagles_first_parts = [eagle[0] for eagle in eagles]
r = [item for item in Result if len(item) > 2 and item[2] in eagles_first_parts]

Answer 3

NB: Not writing most efficient code, but something which follows from your attempts. I am assuming that Result is a List of Lists like:

Result=[['"', '"', 'Fe', '1'],['Hola', 'hola', 'I', '1'],
['como', 'como', 'CS', '0.999289'],
['estas', 'este', 'DD0FP0', '0.97043'],
['Bien', 'bien', 'NP00000', '1'],
['gracias', 'gracia', 'NCFP000', '1'],
['y', 'y', 'CC', '0.999962'],
['tu', 'tu', 'DP2CSS', '1'],
['yo', 'yo', 'PP1CSN00', '1'],
['estoy', 'estar', 'VAIP1S0', '1'],
['bien', 'bien', 'RG', '0.902728'],
['huevo', 'huevo', 'NCMS000', '0.916667'],
['calcio', 'calcio', 'NCMS000', '1'],
['leche', 'leche', 'NCFS000', '1'],
['proteina', 'proteina', 'NCFS000', '1'],
['Francisco', 'francisco', 'NP00000', '1'],
['1999', '1999', 'Z', '1'],
['"', '"', 'Fe', '1']]

Now starting from where you left.

e=[x[0] for x in eagles]

Now, initialize an empty list, r

r=[]

for item in Result:
    for eagle in e:
       if item[2]==eagle:
            r.append(item)
print r

which gives the output:

[['gracias', 'gracia', 'NCFP000', '1'],
['huevo', 'huevo', 'NCMS000', '0.916667'],
['calcio', 'calcio', 'NCMS000', '1']]