df['opposition'].apply(lambda x: x[2:]). Please help me to understand the lambda function and how it works in this case. Thanks in advance.

CodePudding user response：

Few things at play here:

• df[column].apply(f) takes a function f as argument and applies that function to every value in column, returning the new column with modified values.
• lambda x: x[2:] defines a function that takes a value x and returns the slice x[2:]. I.e., when x is a string, it returns x without the first two characters.
• Hence, df['opposition'].apply(lambda x: x[2:]) returns the 'opposition' column modified by removing the first 2 characters from all strings in it.

However, for this particular use case, there is a much better way to do this. You can use .str.slice() to perform the same operation:

df['opposition'].str.slice(start=2)

The methods in .str are specific for columns with string values. See here for more info.

CodePudding user response：

First you can read about lamba function separately and then how it works on a list.

Let's take an example where we have to add 1 to every value present in list. We can do that by iterating each value in a loop and calling the function explicitly.

# Normal iteration method (Approach 1:)
    def add_one_to_every_value(input):
        return input   1
    
    my_list = [1, 2, 3, 4]
    
    res_list = []
    
    for i in my_list:
        res_val = add_one_to_every_value(i)
        res_list.append(res_val)
    
    print(res_list)
    
   # Approach 2: Using map and the function
   res_list2 = list(map(add_one_to_every_value, my_list))
   print(res_list2)

    # Approach 3:
    # The same operation using lambda function. Every value will be iterated 
    # and the corresponding operation (x 1) will be done.
    res_list3 = list(map(lambda x: x 1, my_list))
    print(res_list3)

The same approach is used in pandas apply function.

# Removing first 2 characters in a string is done as follows

test_str = "abcdefg"
print(test_str[2:])

So using this first 2 character removal code inside lambda, we get

df['opposition'].apply(lambda x: x[2:])