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 Whether you're a beginner or brushing up on your skills, these are the real-world questions Python learners ask most about key libraries in data science. Let’s dive in! 🐍 🐼 Pandas: Data Manipulation Made Easy 1. How do I handle missing data in a DataFrame? df.fillna( 0 ) # Replace NaNs with 0 df.dropna() # Remove rows with NaNs df.isna(). sum () # Count missing values per column 2. How can I merge or join two DataFrames? pd.merge(df1, df2, on= 'id' , how= 'inner' ) # inner, left, right, outer 3. What is the difference between loc[] and iloc[] ? loc[] uses labels (e.g., column names) iloc[] uses integer positions df.loc[ 0 , 'name' ] # label-based df.iloc[ 0 , 1 ] # index-based 4. How do I group data and perform aggregation? df.groupby( 'category' )[ 'sales' ]. sum () 5. How can I convert a column to datetime format? df[ 'date' ] = pd.to_datetime(df[ 'date' ]) ...

Here's Python Program for List Duplicates

Here is a program to find the item that occurs most frequently in a data structure. So why to find frequent item? Maybe it is the most purchased item on your shopping site. Perhaps it is the web page that gets hit the most often.

If you are a tester, it could easily be the test that has had the most failures over the last year. Whatever it is, you want an easy way to find the data you need, and Python is here to help you.


Python List duplicates

List frequent item


Here are the two simple lists:

list_1 = [1,2,3,2,3,2] 
list_2 = ['a', 'b', 'a', 'b', 'c']

  • We can't do simple math on the individual items since the second list contains characters. For example, it could contain the words of a book, and you want to find the most commonly used word in the work. 
  • Also, it maybe list of UPC values for commonly purchased items. Whatever it is, all we can guarantee is that the data is probably comparable, in that we can compare one of the items to another. Yet we need to find frequent items.


Python program

Below, you will find a program to find repeated values.

list_1 = [1,2,3,2,3,2]
list_2 = ['a', 'b', 'a', 'b', 'c']

def most_common_brute_force(l):

  # Find the counts of all elements
    dict_of_counts = {}
    for i in l:
        if i in dict_of_counts.keys():
            dict_of_counts[i] = dict_of_counts[i] + 1
        else:
            dict_of_counts[i] = 1

            max_count = -1
            max_value = -1
 
        for k, v in dict_of_counts.items():
            if v > max_count:
                max_count = v
                max_value = k
    return max_value
print(most_common_brute_force(list_1))
print(most_common_brute_force(list_2))

Output

2
   

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