如何计算相同数字在python列表中出现的次数？

Question

I have a list of about 75000 which contain random sequence of +1s and -1s. I want to count how many time +1 has appeared one time and two times and three times and so on. So goes for the -1s.

For example My_list = [1,1,1,-1,-1,1,1,1,-1,-1,1,-1,1...]

expected output would be: (1:1x2,2x0,3x2|-1:1x1,2x2}) in words:+1 repeated once 2 times, twice 0 times, thrice repeated 2 times -1 repeated once 1 time, twice 2 times

Thank you

I am very new to python learning it especially for my trading project. I cant go further than counting the total number occurrence of a given value rather than counting the repeated number of occurrence

Answer 1

Approach

• Using groupby to find runs
• Using Counter to count runs of value +1/-1

Code

from itertools import groupby
from collections import Counter

def run_stats(lst):
    def rle(lst):
        ' Run length encoding of the runs (value, runlength) '
        return [(key, len(list(group))) for key, group in groupby(lst)]
    
    # Count of (value, runlength) pairs
    cnts = Counter(rle(lst))
    
    # Aggregate runs of +1/-1 tuples in lists
    stats = {1:[],    # +1 run pairs
             -1:[]}   # -1 run pairs
    for tup, cnt in cnts.items():
        val, runlength = tup 
        stats[val].append((runlength, cnt))
    
    return stats

Usage

# Test Data
lst = [1,1,1,-1,-1,1,1,1,-1,-1,1,-1,1]

# Generate result
results = run_stats(lst)

# Format Output (using JamieDoombos formatting)
for val in results:
    print(f'{val}:', ','.join(f'{run}x{count}' for run, count in results[val]))

Output

1: 3x2,1x2
-1: 2x2,1x1

Answer 2

I suggest iterating over the list using an index, and for each new item, count the occurrences and store the result in a dict.

The resulting runs dict in this code has the list domain (-1, 1) as keys. Each value in the dict is another dict with the run lengths as keys, and the number of occurrences of the run length as values.

from collections import defaultdict

My_list = [1,1,1,-1,-1,1,1,1,-1,-1,1,-1,1]

# map of values to a map of lengths to the number of occurrences
runs = defaultdict(lambda: defaultdict(int))

list_length = len(My_list)

index = 0
while index < list_length:
    item = My_list[index]
    run_length = 1
    index += 1
    while index < list_length and My_list[index] == item:
        index += 1
        run_length += 1
    runs[item][run_length] += 1


for value, value_runs in runs.items():
    print(f'{value}:', ','.join(f'{run}x{count}' for run, count in value_runs.items()))

Result:

1: 1x2,2x0,3x2
-1: 1x1,2x2,3x0

EDIT: this uses a defaultdict that handles any number of consecutive values and values outside the domain.