How to count the number of times same number has appeared in a list in python?-CodePudding

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

CodePudding user response：

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

CodePudding user response：

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.

CodePudding user response：

You could try this - use groupby with defaultdict(list) to loop and count/group similar 1 or -1 together:

The defaultdict dd will have all 1 or -1 frequency times (spreads) in the final list. Then you choose the way you want to print (format).

L = [1,1,1,-1,-1,1,1,1,-1,-1,1,-1,1,1,1,1,1,-1,-1,-1] # sample input

from itertools import groupby
from collections import defaultdict

        
dd = defaultdict(list)

for k, g in groupby(L, lambda x: x>0):
    if k:  # True
        dd[1].append(len(list(g)))
    else:  # False :: -1
        dd[-1].append(len(list(g)))

        

print(dd)
# defaultdict(<class 'list'>, {1: [3, 3, 1, 5], -1: [2, 2, 1, 3]})