I have a code which gives back 10 lists of numbers.
def my_random_list(l: list):
return sorted(random.sample(list(set(l)), 6))
for _ in range(10):
print(sorted(my_random_list([i for i in range(1, 43)])))
I need to count how many duplicates are there in this 10 lists. How to do it in short and efficient way?
You can use:
import random
from collections import defaultdict
def my_random_list(l: list):
return sorted(random.sample(list(set(l)), 6))
repeated = defaultdict(int)
for _ in range(10):
rl = my_random_list([i for i in range(1, 43)])
for x in rl:
repeated[x] += 1
print(sorted(rl))
repeated = {k:v for k,v in repeated.items() if v > 1}
print(repeated)
# {2: 2, 5: 3, 19: 4, 21: 4, 4: 3, 8: 2, 14: 2, 38: 3, 9: 3, 24: 2, 40: 3, 42: 2, 10: 2, 22: 3, 32: 2, 18: 3, 34: 2, 30: 2, 31: 3}
print(len(repeated.keys())) # how many duplicates
Convert the list to a set, which automatically gets rid of duplicates. Then compare their size:
l = [1,2,3,4,5,6,7,7,6,5,4]
print(len(l) - len(set(l)))
If your intention is to find out the duplicates across the 10 lists, you can try the following -
# Import Counter from collections
In [11]: from collections import Counter
# Your definition of my_random_list
In [12]: def my_random_list(l: list):
...: return sorted(random.sample(list(set(l)), 6))
...:
# Copying your version of creating 10 lists into a lists variable (calling the sorted() here is superfluous in my opinion)
In [13]: lists = [sorted(my_random_list([i for i in range(1, 43)])) for _ in range(10)]
# Count all the entries across all the 10 lists
In [14]: counter = Counter([])
# You can add multiple Counter instances to produce a "merged" Counter
In [15]: for l in lists:
...: counter += Counter(l)
# Find the entries whose value exists more than once
In [16]: duplicates = [k for k,v in counter.items() if v > 1]
# Printing all the duplicate entries across the lists
In [17]: duplicates
Out[17]: [6, 16, 20, 37, 38, 2, 9, 29, 1, 18, 33, 3, 17, 19, 31, 15, 21, 42, 41, 11]
# Length of the duplicate list
In [18]: len(duplicates)
Out[18]: 20
You can read-up on Counter here
A statement of problem is not clear, I assume you want to calculate duplicates in concatenation of these 10 arrays. In this case you could use advantages of numpy.unique:
import random
import numpy as np
collection = [my_random_list(list(range(1, 43))) for i in range(10)]
conc = np.concatenate(collection) # concatenated list
items, cnt = np.unique(conc, return_counts=True) # sorted set of unique items and their counts
output = items[cnt>1] # items that appears more than once
collections.Counter and itertools.chain will be helpful.
import random
source = [i for i in range(1, 43)]
def my_random_list():
return sorted(random.sample(source, 6))
random_lists = [my_random_list() for _ in range(10)]
print(random_lists)
Here are 10 random lists(6 length for each).
>>> [[2, 4, 10, 18, 20, 30], [4, 12, 13, 19, 21, 27], [10, 11, 18, 26, 32, 33], [4, 11, 12, 17, 38, 42], [12, 22, 28, 38, 40, 41], [2, 11, 22, 30, 35, 36], [4, 6, 22, 24, 32, 34], [1, 3, 5, 25, 31, 33], [25, 29, 31, 32, 33, 35], [12, 16, 28, 31, 37, 41]]
Then you can count it.
from collections import Counter
from itertools import chain
counter = Counter(chain(*random_lists))
print(counter)
>>> Counter({4: 4, 12: 4, 11: 3, 32: 3, 33: 3, 22: 3, 31: 3, 2: 2, 10: 2, 18: 2, 30: 2, 38: 2, 28: 2, 41: 2, 35: 2, 25: 2, 20: 1, 13: 1, 19: 1, 21: 1, 27: 1, 26: 1, 17: 1, 42: 1, 40: 1, 36: 1, 6: 1, 24: 1, 34: 1, 1: 1, 3: 1, 5: 1, 29: 1, 16: 1, 37: 1})
And filter the counter with comprehension.
results = [k for k, v in counter.items() if v >= 2]
print(results)
>>> [2, 4, 10, 18, 30, 12, 11, 32, 33, 38, 22, 28, 41, 35, 25, 31]
collections.Counter. Counter is bag data structure of python. If you put all your lists into a Counter, you can get what elements are duplicate(greater than 2) and how many overlap.