Python根据给定的键合并2个列表

Question

I am dealing with a problem, here is the input

list1 = ['A', 'U', 'C', 'C', 'A']
list2 = ['12', '14']
key = {'A12':'*', 'C14':'#'}

the output like this:

output1 = [['A12', 'U', 'C14', 'C', 'A'], ['A12', 'U', 'C', 'C14', 'A'],['A', 'U', 'C14', 'C', 'A12'], ['A', 'U', 'C', 'C14', 'A12']]

and convert to

output2 = [['*', 'U', '#', 'C', 'A'], ['*', 'U', 'C', '#', 'A'],['A', 'U', '#', 'C', '*'], ['A', 'U', 'C', '#', '*']]

I am using Python2.7 to solve this problem, but I have not figured it out yet... Any answer or suggestion will be appreciated!

Here is my code:

list1 = ['A', 'U', 'C', 'C', 'A']
list2 = ['12', '14']
key = {'A12':'*', 'C14':'#'}

list3 = ['12', '14', '0', '0','0'] #build by myself
combo = list(set(itertools.permutations(list3, len(list3))))
list_combo = []
for each_list in combo:
new_list = []
for i in xrange(len(list1)):
    if list1[i]+each_list[i] in key:
        new_list.append(key[list1[i]+each_list[i]])
    else:
        new_list.append(list1[i])
    list_combo.append(new_list)
print list_combo

There are some extra lists in the output, and if list2 or list3 is too big, it will take a lot of time to run itertools.permutations, so I am seeking another way to solve this problem.

Answer 1

Ok this is a little long, so bear with me. The first step is constructing a dictionary to convert from letters to letters with numbers, ie A to A12 , etc.

replacements = dict((k[0],k) for k in key.keys())
# replacements is equal to {'A': 'A12', 'C': 'C14'}

This makes things much simpler later. The next step is building a list of all the indicies that need replacing and breaking those into sublists for each type of replacement.

indicies = [[i for i,x in enumerate(list1) if x == k] for k in replacements.keys()]
# indicies is equal to [[0, 4], [2, 3]]

Finally, we use itertools.product on the list of indices that need replacing to get each possible grouping, and then add them to the necessary output lists:

output1 = []
output2 = []
for group in itertools.product(*indicies):
    l = []
    l2 = []
    for i in range(len(list1)):
        l.append(list1[i] if i not in group else replacements[list1[i]])
        l2.append(list1[i] if i not in group else key[replacements[list1[i]]])
    output1.append(l)
    output2.append(l2)
print output1
print output2

This gives us our desired answers of:

[['A12', 'U', 'C14', 'C', 'A'], ['A12', 'U', 'C', 'C14', 'A'], ['A', 'U', 'C14', 'C', 'A12'], ['A', 'U', 'C', 'C14', 'A12']]
[['*', 'U', '#', 'C', 'A'], ['*', 'U', 'C', '#', 'A'], ['A', 'U', '#', 'C', '*'], ['A', 'U', 'C', '#', '*']]

One major difference between this code and the code that you are running is that my code only runs for exactly as many iterations as it needs to, so in the case of your sample data set, 4 times. Your code seems like it runs at least at least certain parts n! times, where n is the size of list1 , and is equal to 120 for your sample data set. This is still likely to run for a while on extremely large data sets (as is the nature of this sort of problem), but will only scale up with the number of replacements it has to do, rather than the size of the data set overall.