从Python中的元组列表中查找相等的值

Question

After searching a lot without success I need help.

I have a list of list of tuples. Each list inside the list of list represent certain numbers of formulas in my system. Any element in this list is a tuple that represent the type of the element (variable, parameter, constant, an operation...) and the name of the element. For example for the formulas x1+x2+A1 , x1-x3 and sin(x2)+A1 we'll have:

[
[('VAR', 'x1'), ('PLUS', '+'), ('VAR', 'x2'), ('PLUS', '+'), ('PAR', 'A1')],
[('VAR', 'x1'), ('LESS', '-'), ('VAR', 'x3')],
[('SIN', 'sin'), ('VAR', 'x2'), ('PLUS', '+'), ('PAR', 'A1')]
]

I'm trying to determine in which formula each variable appear. In the example above I have that x1 variable is on 1 and 2 formula, x2 variable is on 1 and 3 formula and x3 in 2 formula, so my output will be something like:

[
['x1', 1, 2],
['x2', 1, 3],
['x3', 2],
]

At the moment I have very inefficient code that doesn't work at all, but here it is:

cont = 0
for subL1 in L:
    for subL2 in L:
        if len(subL1) != 1 and len(subL2) != 1:
            if subL1 != subL2 and subL2:
                for x,y in subL1:
                    for z,t in subL2:
                        if (    x == 'VAR'
                            and z == 'VAR'
                            and y == t
                            ):
                            print "Variable", y , "repeated"
        else:
            print "list with 1 lenght\n"
    subL1.pop(0)
cont = cont + 1

Answer 1

You could use a collections.defaultdict to store the formulas (actually the indices inside your list of lists) for each variable:

from collections import defaultdict

dd = defaultdict(set)              # use a set as factory so we don't keep duplicates
for idx, subl in enumerate(l, 1):  # iterate over the sublists with index starting at 1
    for subt in subl:              # iterate over each tuple in each sublist
        label, val = subt          # unpack the tuple
        if label == 'VAR':         # if it's a VAR save the index in the defaultdict
            dd[val].add(idx)

For example with:

l = [[('VAR', 'x1'), ('PLUS', '+'), ('VAR', 'x2'), ('PLUS', '+'), ('PAR', 'A1')],
     [('VAR', 'x1'), ('LESS', '-'), ('VAR', 'x3')],
     [('SIN', 'sin'), ('VAR', 'x2'), ('PLUS', '+'), ('PAR', 'A1')]
    ]

It gives:

print(dd)
# defaultdict(set, {'x1': {1, 2}, 'x2': {1, 3}, 'x3': {2}})

To get your desired output you only need to convert that to a list again, for example (python-3.x only):

>>> [[name, *sorted(formulas)] for name, formulas in sorted(dd.items())]
[['x1', 1, 2], ['x2', 1, 3], ['x3', 2]]

Answer 2

formula = [
[('VAR', 'x1'), ('PLUS', '+'), ('VAR', 'x2'), ('PLUS', '+'), ('PAR', 'A1')],
[('VAR', 'x1'), ('LESS', '-'), ('VAR', 'x3')],
[('SIN', 'sin'), ('VAR', 'x2'), ('PLUS', '+'), ('PAR', 'A1')]
]

variables = collections.defaultdict(set)
for line_no, line in enumerate(formula):
    for typ, value in line:
        if typ == 'VAR':
            variables[value].add(line_no)

 variables 

defaultdict(set, {'x1': {0, 1}, 'x2': {0, 2}, 'x3': {1}})