Geek asks and answers

The total number of similar items that meet the criteria of uniqueness

I have a list rodsthat consists of tuples lengthand position. positionalways unique to the given length. I want to find the most frequent length of the bar, and then the total number of occurrences of all unique (on position) neighboring bars (including the most frequent). Broken:

  • First I want to find the most common core length.
  • Then I want to include all other members that have an adjacent lengthmember by some criterion (+ -1 in this example), BUT ONLY IF UNIQUE THE UNIQUE POSITION - ALREADY AN ACCOUNT FOR (or "most often") member in the original group or "new core "added to this group by completing adjacent criteria).
  • And find this new common frequency.

I can accomplish this as follows, sorting and using sets, but maybe there is a better solution:

import itertools
#tuples of (length, position)
rods = [(18, 21), (17, 2), (15, 3), (14, 21), (14, 5), (13, 6), (13, 7),
        (13, 8), (13, 9), (13, 10), (13, 11), (13, 12), (13, 13), (13, 14),
        (13, 15), (13, 16), (13, 17), (13, 18), (13, 19), (13, 20), (13, 21),
        (13, 22), (13, 23), (13, 24), (13, 25), (13, 26), (12, 5), (12, 21),
        (12, 2)]

lengths = [length for length, position in rods]

#gives tuples of lengths and their frequencies:
length_freq = (sorted([(k,len(list(j))) for k,j in itertools.groupby(sorted(lengths))],
               key=lambda x: x[1],reverse=1))
best_length = length_freq[0][0]

#cumulative frequency of rods near best_length, with unique position:
tally = (len(set((best_length,v) for j,v in rods 
         if best_length - 1 <= j <=best_length + 1)))

print length_freq
#output:
#[(13, 21), (12, 3), (14, 2), (15, 1), (17, 1), (18, 1)]
print tally
#output:
#23

Note 23- The correct answer is for this test data. Since both rods with length= 14are located at points also occupied by rods with length=15(position 21and 5). And there is also an overlap in position=21for lengths 13 and 12.

+3
source share
2 answers

, , . - . , groupby Counter, , defaultdict, . groupby - ; , , , , , .

Nolen Royalty defaultdict, Counter , . . O (n); , O (n log n), .

import collections

#tuples of (length, position)
rods = [(18, 21), (17, 2), (15, 3), (14, 21), (14, 5), (13, 6), (13, 7),
        (13, 8), (13, 9), (13, 10), (13, 11), (13, 12), (13, 13), (13, 14),
        (13, 15), (13, 16), (13, 17), (13, 18), (13, 19), (13, 20), (13, 21),
        (13, 22), (13, 23), (13, 24), (13, 25), (13, 26), (12, 5), (12, 21),
        (12, 2)]

lengths = (length for length, position in rods)
length_freq = collections.Counter(lengths)
((best_length, _),) = length_freq.most_common(1)
print best_length

#cumulative frequency of rods near best_length, with unique position:
rod_filter = ((l, p) for l, p in rods if best_length - 1 <= l <= best_length + 1)
tally = len(set((best_length, p) for l, p in rod_filter))

print length_freq
print tally

Counter, , . :

length_freq = collections.Counter(lengths)
((best_length, _),) = length_freq.most_common(1)

:

length_freq = collections.defaultdict(int)
for l in lengths:
    length_freq[l] += 1
best_length = max(length_freq, key=length_freq.get)

, ; .

+2

, :

>>> from collections import defaultdict
>>> rods = [(18, 21), (17, 2), (15, 3), (14, 21), (14, 5), (13, 6), (13, 7),
...         (13, 8), (13, 9), (13, 10), (13, 11), (13, 12), (13, 13), (13, 14),
...         (13, 15), (13, 16), (13, 17), (13, 18), (13, 19), (13, 20), (13, 21),
...         (13, 22), (13, 23), (13, 24), (13, 25), (13, 26), (12, 5), (12, 21),
...         (12, 2)]
>>> neighbor_cutoff = 1
>>> length_to_count = defaultdict(int)
>>> neighbors_for_length = defaultdict(set)
>>> for rod in rods:
...     length_to_count[rod[0]] += 1
...     neighbors_for_length[rod[0]].add(rod[1])
...     for i in range(1, neighbor_cutoff+1):
...         neighbors_for_length[rod[0]-i].add(rod[1])
...         neighbors_for_length[rod[0]+i].add(rod[1])
... 
>>> sorted([(length, length_to_count[length]) for length in length_to_count], key=lambda x: x[1], reverse=True)
[(13, 21), (12, 3), (14, 2), (15, 1), (17, 1), (18, 1)]
>>> [(length, len(neighbors_for_length[length])) for length in neighbors_for_length]
[(11, 3), (12, 23), (13, 23), (14, 23), (15, 3), (16, 2), (17, 2), (18, 2), (19, 1)]
>>> sorted(_, key=lambda x: x[1], reverse=True)
[(12, 23), (13, 23), (14, 23), (11, 3), (15, 3), (16, 2), (17, 2), (18, 2), (19, 1)]
>>> neighbors_for_length
defaultdict(<type 'set'>, {11: set([2, 5, 21]), 12: set([2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26]), 
13: set([2, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26]), 
14: set([3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26]),
15: set([3, 21, 5]), 16: set([2, 3]), 17: set([2, 21]), 18: set([2, 21]), 19: set([21])})
+1

More articles:


All Articles