Python – Sort by a particular digit count in elements
Last Updated :
08 Mar, 2023
Given a list of elements, sort by K digit in each element.
Examples:
Input : test_list = [4322, 2122, 123, 1344], K = 2
Output : [1344, 123, 4322, 2122]
Explanation : 0 < 1 < 2 < 3, sorted by count of 2 in each element.
Input : test_list = [4322, 2122, 1344], K = 2
Output : [1344, 4322, 2122]
Explanation : 0 < 2 < 3, sorted by count of 2 in each element.
Method #1: Using list comprehension + str() + count()
In this, we perform the task of sorting using sort(), and the task of finding frequency is done using count().
Python3
def count_dig(ele):
return str(ele).count(str(K))
test_list = [4322, 2122, 123, 1344]
print("The original list is : " + str(test_list))
K = 2
test_list.sort(key = count_dig)
print("Sorted list : " + str(test_list))
|
Output
The original list is : [4322, 2122, 123, 1344]
Sorted list : [1344, 123, 4322, 2122]
Time Complexity: O(n*nlogn)
Auxiliary Space: O(1)
Method #2 : Using sorted() + str() + count() + lambda
In this, we perform the task of performing sort using sorted(), and use the lambda function to get sort logic rather than the external function.
Python3
test_list = [4322, 2122, 123, 1344]
print("The original list is : " + str(test_list))
K = 2
res = sorted(test_list, key = lambda ele : str(ele).count(str(K)))
print("Sorted list : " + str(res))
|
Output
The original list is : [4322, 2122, 123, 1344]
Sorted list : [1344, 123, 4322, 2122]
Time Complexity: O(n*logn), where n is the length of the input list. This is because we’re using the built-in sorted() function which has a time complexity of O(nlogn) in the worst case.
Auxiliary Space: O(1), as we’re not using any additional space other than the input list itself.
Method #3 : Using count(),sort(),extend() and index() methods
Python3
test_list = [4322, 2122, 123, 1344]
print("The original list is : " + str(test_list))
K = 2
x=[]
for i in test_list:
x.append(str(i).count(str(K)))
y=[]
y.extend(x)
y.sort()
res=[]
for i in y:
res.append(test_list[x.index(i)])
print("Sorted list : " + str(res))
|
Output
The original list is : [4322, 2122, 123, 1344]
Sorted list : [1344, 123, 4322, 2122]
Method 4: using operator.countOf() method
Python3
import operator as op
test_list = [4322, 2122, 123, 1344]
print("The original list is : " + str(test_list))
K = 2
res = sorted(test_list, key = lambda ele : op.countOf(str(ele),str(K)))
print("Sorted list : " + str(res))
|
Output
The original list is : [4322, 2122, 123, 1344]
Sorted list : [1344, 123, 4322, 2122]
Time Complexity: O(NLogN)
Auxiliary Space : O(N)
Method 5:Using the itertools.groupby() function
Python3
import itertools
test_list = [4322, 2122, 123, 1344]
K = 2
test_list.sort(key = lambda x: str(x).count(str(K)))
sorted_list = [list(g) for k, g in itertools.groupby(test_list, key = lambda x: str(x).count(str(K)))]
print(sorted_list)
|
Output
[[1344], [123], [4322], [2122]]
Time Complexity: O(NLogN)
Auxiliary Space : O(N)
Method 6:Using a dictionary :
Algorithm:
1.Create an empty dictionary to store the numbers with their frequency of digit ‘K’.
2.Iterate through the input list and for each number in the list:
a. Convert the number to string.
b. Count the frequency of digit ‘K’ in the string.
c. Add the number to the dictionary with its frequency of digit ‘K’ as key.
3.Sort the dictionary based on keys (i.e. frequency of digit ‘K’).
4.Create a list of lists where each sub-list contains all the numbers having the same frequency of digit ‘K’.
5.Return the list of lists.
Python3
def group_numbers_by_frequency_using_dict(test_list, K):
dict_freq = {}
for num in test_list:
num_str = str(num)
freq = num_str.count(str(K))
if freq in dict_freq:
dict_freq[freq].append(num)
else:
dict_freq[freq] = [num]
dict_freq_sorted = dict(sorted(dict_freq.items()))
res = [dict_freq_sorted[key] for key in dict_freq_sorted]
return res
test_list = [4322, 2122, 123, 1344]
K = 2
print(group_numbers_by_frequency_using_dict(test_list, K))
|
Output
[[1344], [123], [4322], [2122]]
Time Complexity: O(n * m * log m), where n is the number of elements in the input list and m is the maximum number of digits in an element of the input list.
Auxiliary Space: O(n), where n is the number of elements in the input list.
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