In this article, we are going to see how we will return a boolean array which is True where the string element in the array ends with a suffix in Python.
numpy.char.endswith()
numpy.char.endswith() return True if the elements end with the given substring otherwise it will return False.
Syntax : np.char.endswith(input_numpy_array,’substring’)
Parameters:
- input_numpy_array refers to the numpy array with strings
- substring is compared with all elements present in an array
Return: Return the boolean array which includes “True” if a substring is present as a suffix and “False” if a substring is not present as a suffix.
Example 1:
In this example, we are creating a NumPy array with 5 strings and checking the elements’ ends with ‘ks’.
Python3
# import numpyimport numpy as np# Create 1D array of strings.a = np.array(['hello', 'welcome to', 'neveropen', 'for', 'neveropen'])# check the strings in an above array# ends with substring - 'ks'# and stored in a variable "gfg".Data = np.char.endswith(a, 'ks')# Print boolean arrayprint(Data) |
Output:
[False False True False True]
Example 2:
In this example, we are creating a NumPy array with 5 strings and checking the element’s ends with ‘o’.
Python3
# import numpyimport numpy as np# Create 1D array of strings.a = np.array(['hello', 'welcome to', 'neveropen', 'for', 'neveropen'])# check the strings in an above array# ends with substring - 'o'# and stored in a variable "gfg".Data = np.char.endswith(a, 'o')# Print boolean arrayprint(Data) |
Output:
[ True True False False False]
Time complexity: O(N), where N is the number of strings in the array “a”.
Space complexity : O(N), as we are storing the result in a boolean array “Data” with the same number of elements as the number of strings in the array “a”.