I have a list in python that looks like this:
data = ['12,34,212,90,1,8','901,2,1,8,44,1,1','77,32,11,230,894','78,23,45,89,12,20']
How do I count the number of occurrences for each number?
Output should be something like...
12:x
34:x
212:x
Where x is the number of times the value is found in the whole list
I've tried the following but not working as intended.
d = {}
#[ d.update( {i:d.get(i, 0)+1} ) for i in data ]
data = ['12,34,212,90,1,8', '901,2,1,8,44,1,1', '77,32,11,230,894',
'78,23,45,89,12,20']
d = {}
for item in data:
for i in item.split(','):
d[i] = d.get(i, 0) + 1
for i, j in d.items():
print(f'{i}:{j}')
explanation : You first iterate over the data list, then you split the strings to get the numbers, then d[i] = d.get(i, 0) + 1 will do the counting. finally I iterated over the result dictionary and prints the output you want with f-string.
Convert the original list to a list of strings where the original strings are split.
data = ['12,34,212,90,1,8','901,2,1,8,44,1,1','77,32,11,230,894','78,23,45,89,12,20']
t = []
{t.extend(e.split(',')) for e in data}
print(t)
The resulting list would look like this:
['12', '34', '212', '90', '1', '8', '901', '2', '1', '8', '44', '1', '1', '77', '32', '11', '230', '894', '78', '23', '45', '89', '12', '20']
Then you simply use the count() method of the list. For instance:
t.count('1')
Will give you:
4
First, convert the list of string into a list containing just the values.
res = []
for i in data:
res += i.split(',')
Then we can use python's Counter to count each value:
from collections import Counter
res = []
data = ['12,34,212,90,1,8','901,2,1,8,44,1,1','77,32,11,230,894','78,23,45,89,12,20']
for i in data:
res += i.split(',')
res = Counter(res)
for key in res:
print(f"{key} exist {res[key]} times")
Output:
12 exist 2 times
34 exist 1 times
212 exist 1 times
90 exist 1 times
1 exist 4 times
8 exist 2 times
901 exist 1 times
2 exist 1 times
44 exist 1 times
77 exist 1 times
32 exist 1 times
11 exist 1 times
230 exist 1 times
894 exist 1 times
78 exist 1 times
23 exist 1 times
45 exist 1 times
89 exist 1 times
20 exist 1 times
using Counter and join
# your code goes here
data = ['12,34,212,90,1,8','901,2,1,8,44,1,1','77,32,11,230,894','78,23,45,89,12,20']
from collections import Counter
result = Counter(map(int, ','.join(data).split(',')))
print(result)
# output
# Counter({1: 4, 12: 2, 8: 2, 34: 1, 212: 1, 90: 1, 901: 1, 2: 1, 44: 1, 77: 1, 32: 1, 11: 1, 230: 1, 894: 1, 78: 1, 23: 1, 45: 1, 89: 1, 20: 1})
Let's break down in two steps.
1. Retrieve the numbers
You have a list of strings and each string has numbers separated by commas that need to be parsed.
>>> numbers = [int(x) for s in data for x in s.split(',')]
>>> numbers
[12, 34, 212, 90, 1, 8, 901, 2, 1, 8, 44, 1, 1, 77, 32, 11, 230, 894, 78, 23, 45, 89, 12, 20]
tip: the correct order of the for statements when flatting in a comprehension is the same as if you would iterate using two nested fors.
2. Count!
>>> from collections import Counter
>>> Counter(numbers)
Counter({1: 4, 12: 2, 8: 2, 34: 1, 212: 1, 90: 1, 901: 1, 2: 1, 44: 1, 77: 1, 32: 1, 11: 1, 230: 1, 894: 1, 78: 1, 23: 1, 45: 1, 89: 1, 20: 1})
d = dict()
for i in data:
temp = i.split(",")
for num in temp:
n = int(num)
if n not in d:
d[n] = 1
else:
d[n] += 1
