Count number of occurences for each position in nested list

Question

I have a list X with numbers from 1-10 , and a function that uses this list to create a new list with random values from 1-10. I want to call this function 100 times and count the number of times the same value occurs on the same position in the nested list.

I've created the function lists to achieve this. My code returns the correct answer, but I'm thinking there has to be a simpler way to write it:

from collections import Counter import random

X = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

def randomX(l):
    randomX = [random.choice(X) for i in range(len(data))]
    return randomX

def lists(times):
    lists = [randomX(X) for i in range(times)]
    x1 = Counter(sublist[0] for sublist in lists)
    x2 = Counter(sublist[1] for sublist in lists)
    x3 = Counter(sublist[2] for sublist in lists)
    x4 = Counter(sublist[3] for sublist in lists)
    x5 = Counter(sublist[4] for sublist in lists)
    x6 = Counter(sublist[5] for sublist in lists)
    x7 = Counter(sublist[6] for sublist in lists)
    x8 = Counter(sublist[7] for sublist in lists)
    x9 = Counter(sublist[8] for sublist in lists)
    x10 = Counter(sublist[9] for sublist in lists)
    return x1, x2, x3, x4, x5, x6, x7, x8, x9, x10

print(lists(100))

This prints a result that looks like this, which is kind of the thing I want, except that I'm going to sort it. However, it seems unnecessary to write pretty much the same thing 10 times in a row, but list slicing won't work with counter.

(Counter({7: 14, 8: 12, 6: 12, 3: 11, 9: 11, 1: 10, 10: 9, 4: 9, 5: 7, 2: 5}),
 Counter({5: 16, 3: 14, 7: 11, 4: 11, 10: 9, 1: 9, 6: 8, 8: 8, 2: 7, 9: 7}),
 Counter({3: 14, 2: 14, 7: 13, 8: 13, 4: 10, 6: 10, 5: 10, 1: 8, 9: 5, 10: 3}),
 Counter({3: 15, 6: 15, 8: 12, 7: 11, 1: 11, 4: 11, 2: 10, 9: 7, 5: 5, 10: 3}),
 Counter({8: 20, 3: 15, 6: 13, 4: 11, 7: 10, 10: 10, 2: 7, 1: 7, 5: 4, 9: 3}),
 Counter({9: 15, 6: 13, 10: 12, 4: 11, 2: 10, 8: 9, 3: 8, 5: 8, 1: 8, 7: 6}),
 Counter({6: 17, 7: 13, 9: 11, 2: 11, 8: 11, 5: 10, 3: 8, 10: 8, 4: 6, 1: 5}),
 Counter({6: 20, 5: 11, 10: 11, 1: 11, 9: 10, 2: 9, 4: 8, 3: 8, 7: 7, 8: 5}),
 Counter({8: 13, 10: 13, 4: 13, 5: 11, 9: 11, 1: 10, 3: 9, 2: 8, 7: 6, 6: 6}),
 Counter({10: 14, 8: 13, 1: 10, 2: 10, 3: 10, 5: 10, 7: 9, 4: 9, 9: 9, 6: 6}))

Does anyone have any suggestions as to how I can simplify this code?

Answer 1

You could transpose your created list to columns using zip() and count those.

Small example for what zip() does:

d = [ [1,2,3,4], [11,12,13,14], [111,112,113,114]]

print(list(zip(*d)))  

Output:

# transposed - each tuple equals one column now
[(1, 11, 111), (2, 12, 112), (3, 13, 113), (4, 14, 114)]

---

Your code using zip() and a generator:

from collections import Counter
import random

random.seed(42)

def countEm(data, times):
    datalen = len(data)
    numbers = [ random.choices(data, k=datalen) for _ in range(times)]
    for col in zip(*numbers):
        yield Counter(col) 

X = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

# generators can only be consumed once, use  result = list(countEm(X,100)) if you
# want to reuse the result beside printing... 

for count in countEm(X,100):
    print(count) 

Output:

Counter({9: 18, 7: 13, 10: 13, 5: 12, 3: 11, 8: 11, 1: 8, 4: 6, 6: 4, 2: 4})
Counter({3: 16, 7: 14, 6: 13, 8: 11, 10: 10, 1: 9, 4: 9, 5: 8, 9: 6, 2: 4})
Counter({6: 13, 7: 13, 3: 12, 9: 12, 2: 11, 5: 10, 10: 9, 8: 8, 4: 7, 1: 5})
Counter({10: 17, 2: 15, 6: 12, 3: 9, 9: 9, 5: 9, 8: 9, 7: 8, 1: 7, 4: 5})
Counter({9: 14, 10: 11, 4: 11, 7: 10, 1: 10, 5: 10, 3: 9, 6: 9, 8: 8, 2: 8})
Counter({7: 17, 3: 12, 8: 11, 9: 11, 10: 10, 4: 9, 2: 9, 5: 8, 1: 7, 6: 6})
Counter({3: 13, 2: 13, 5: 13, 1: 11, 10: 11, 6: 10, 8: 9, 7: 8, 9: 7, 4: 5})
Counter({6: 15, 3: 13, 5: 13, 1: 12, 10: 12, 7: 11, 8: 8, 4: 7, 9: 5, 2: 4})
Counter({5: 15, 1: 12, 8: 12, 7: 10, 2: 10, 6: 10, 9: 9, 3: 9, 4: 8, 10: 5})
Counter({3: 13, 9: 13, 4: 12, 10: 10, 2: 10, 1: 9, 8: 9, 7: 8, 6: 8, 5: 8})

---

Remarks:

Our random numbers will not match even if you use the same random.seed(42) . Using n times random.choice() changes the internal random Mersenne_Twister -state different from using random.choices(.., k=n) - if you switched to random.choices you will get identical outputs:

from collections import Counter
import random

random.seed(42)
X = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

def randomX(l):
    lenl=len(l)
    return random.choices(l,k=lenl) 

def lists(times):
    lists = [randomX(X) for i in range(times)]
    x1 = Counter(sublist[0] for sublist in lists)
    x2 = Counter(sublist[1] for sublist in lists)
    x3 = Counter(sublist[2] for sublist in lists)
    x4 = Counter(sublist[3] for sublist in lists)
    x5 = Counter(sublist[4] for sublist in lists)
    x6 = Counter(sublist[5] for sublist in lists)
    x7 = Counter(sublist[6] for sublist in lists)
    x8 = Counter(sublist[7] for sublist in lists)
    x9 = Counter(sublist[8] for sublist in lists)
    x10 = Counter(sublist[9] for sublist in lists)
    return x1, x2, x3, x4, x5, x6, x7, x8, x9, x10

print(*lists(100), sep="\n")

Output now:

Counter({9: 18, 7: 13, 10: 13, 5: 12, 3: 11, 8: 11, 1: 8, 4: 6, 6: 4, 2: 4})
Counter({3: 16, 7: 14, 6: 13, 8: 11, 10: 10, 1: 9, 4: 9, 5: 8, 9: 6, 2: 4})
Counter({6: 13, 7: 13, 3: 12, 9: 12, 2: 11, 5: 10, 10: 9, 8: 8, 4: 7, 1: 5})
Counter({10: 17, 2: 15, 6: 12, 3: 9, 9: 9, 5: 9, 8: 9, 7: 8, 1: 7, 4: 5})
Counter({9: 14, 10: 11, 4: 11, 7: 10, 1: 10, 5: 10, 3: 9, 6: 9, 8: 8, 2: 8})
Counter({7: 17, 3: 12, 8: 11, 9: 11, 10: 10, 4: 9, 2: 9, 5: 8, 1: 7, 6: 6})
Counter({3: 13, 2: 13, 5: 13, 1: 11, 10: 11, 6: 10, 8: 9, 7: 8, 9: 7, 4: 5})
Counter({6: 15, 3: 13, 5: 13, 1: 12, 10: 12, 7: 11, 8: 8, 4: 7, 9: 5, 2: 4})
Counter({5: 15, 1: 12, 8: 12, 7: 10, 2: 10, 6: 10, 9: 9, 3: 9, 4: 8, 10: 5})
Counter({3: 13, 9: 13, 4: 12, 10: 10, 2: 10, 1: 9, 8: 9, 7: 8, 6: 8, 5: 8})