CODEWITHSUNDEEP
pythonarrayslistloops
NorthFred
NorthFred

Reputation: 490

Going through 2 lists with array data

This one is causing me a headache, and I am having trouble to find a solution with a for-loop.

Basically, my data looks like this:

short_list = [ [1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12] ]
long_list  = [ [1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [6, 7, 8, 9, 10], [9, 10, 11, 12, 13] ]

I would need to know how many times each number from each row in the short_list appears in each row of the long_list, and the comparison is NOT needed when both list indices are the same, because they come from the same data set.

Example: I need to know the occurrence of each number in [1, 2, 3] in the long_list rows [2, 3, 4, 5, 6], [6, 7, 8, 9, 10] and [9, 10, 11, 12, 13]. And then continue with the next data row in short_list, etc.

Upvotes: 5

Views: 93

Answers (5)

RoadRunner
RoadRunner

Reputation: 26315

Possible Approach:

  • Loop over each list in short_list.
  • Flatten every list in long_list that is not the same index as the current list, and convert it to a set.
  • Create a collections.Counter() to store the counts for each element in short list that appears in the flattened list.

Demo:

from collections import Counter
from itertools import chain

short_list = [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]]
long_list  = [[1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [6, 7, 8, 9, 10], [9, 10, 11, 12, 13]]

for i, short_lst in enumerate(short_list):
    to_check = set(chain.from_iterable(long_list[:i] + long_list[i+1:]))
    print(Counter(x for x in short_lst if x in to_check))

Output:

Counter({2: 1, 3: 1})
Counter({4: 1, 5: 1, 6: 1})
Counter({9: 1})
Counter({10: 1})

Upvotes: 2

jpp
jpp

Reputation: 164623

This is a brute-force solution. I've amended the input data to make the results more interesting:

from collections import Counter
from toolz import concat

short_list = [ [1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12] ]
long_list  = [ [1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [6, 7, 8, 9, 10], [2, 3, 11, 12, 13] ]

for idx, i in enumerate(short_list):
    long_list_filtered = (x for x in concat(long_list[:idx] + long_list[idx+1:]) if x in set(i)))
    print(idx, Counter(long_list_filtered))

# 0 Counter({2: 2, 3: 2})
# 1 Counter({4: 1, 5: 1, 6: 1})
# 2 Counter()
# 3 Counter({10: 1})

Upvotes: 2

kszl
kszl

Reputation: 1213

for L1 in short_list:
  for L2 in long_list:
    if not set(L1).issubset(set(L2)):
      for x in L1:
        print("{} has {} occurrences in {}".format(x, L2.count(x), L2))

Upvotes: 1

mhawke
mhawke

Reputation: 87064

Here's one way to do it. It's straight off the top of my head, so there is probably a much better way to do it.

from collections import defaultdict

short_list = [ [1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12] ]
long_list  = [ [1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [6, 7, 8, 9, 10], [9, 10, 11, 12, 13] ]

occurrences = defaultdict(int)

for i, sl in enumerate(short_list):
    for j, ll in enumerate(long_list):
        if i != j:
            for n in sl:
                occurrences[n] += ll.count(n)

>>> occurrences
defaultdict(<class 'int'>, {1: 0, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 0, 8: 0, 9: 1, 10: 1, 11: 0, 12: 0})

Note that enumerate() is used to provide indices while iterating. The indices are compared to ensure that sub-lists at the same relative position are not compared.

The result is a dictionary keyed by items from the short list with the values being the total count of that item in the long list sans the sublist with the same index.

Upvotes: 3

Shiva Kishore
Shiva Kishore

Reputation: 1701

short_list = [ [1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12] ]
long_list  = [ [1, 2, 3, 4, 5], [2, 3, 4, 5, 6], [6, 7, 8, 9, 10], [9, 10, 11, 12, 13] ]
occ = []
for si in short_list:
    occi = []
    for i, e in enumerate(si):
        count = 0
        for li in long_list:
            for j, e1 in enumerate(li):
                if i == j:
                    continue
                elif e == e1:
                    count += 1
        occi.append(count)
    occ.append(occi)
print occ

This should work, Happy coding :)

Upvotes: 0

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