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Python – Check if a list is empty or not
  • Last Updated : 13 Sep, 2020

In this article, we will learn How to check if given list is Empty or not. There are various ways in which a list can be checked in Python, but all are not appropriate or in the terms of Python, “pythonic”, to implement.

  1. Let’s see how we can check a list is empty or not, in a less pythonic way. We should avoid this way of explicitly checking for a sequence or list




    # Python code to check for empty list
    # Explicit way
    def Enquiry(lis1):
        if len(lis1) == 0:
            return 0
        else:
            return 1
              
    # Driver Code
    lis1 = []
    if Enquiry(lis1):
        print ("The list is not empty")
    else:
        print("Empty List")

    Output:

    Empty List
    
  2. Now let’s see a more pythonic way to check for an empty list. This method of check is an implicit way of checking and more preferable than the previous one.




    # Python code to check for empty list
    # IMPLICIT way or Pythonic way
    def Enquiry(lis1):
        if not lis1:
            return 1
        else:
            return 0
              
    # Driver Code
    lis1 = []
    if Enquiry(lis1):
        print ("The list is Empty")
    else:
        print ("The list is not empty")

    Output:

    The list is Empty

Numpythonic way

  1. The previous methods that we used in normal Python don’t work for the Numpythonic way. Other methods that work fine for lists or other standard containers fail for numpy arrays. This way fails with numpy arrays because numpy tries to cast the array to an array of bools and if this tries to evaluate all of those bools at once for some kind of aggregate truth value, it fails so we get a ValueError.






    # Numpythonic way with the previous method
    # Returns ValueError
    import numpy
    def Enquiry(lis1):
        return(numpy.array(lis1))
          
    # Driver Code
    lis1 = [0, 1]
    if Enquiry(lis1):
        print("Not Empty")
    else:
        print("Empty")

    Output:

    None

    Error:

    Traceback (most recent call last):
      File "/home/2d237324bb5211d7216c521441a750e9.py", line 7, in 
        if Enquiry(lis1):
    ValueError: The truth value of an array with more than 
    one element is ambiguous. Use a.any() or a.all()
  2. In the next example, we will see that even if the list is Not Empty, the output will show Empty. If the list contains one 0, then the if statement will incorrectly result.




    # Numpythonic way with the previous method
    # Returns wrong result
    import numpy
    def Enquiry(lis1):
        return(numpy.array(lis1))
          
    # Driver Code
    lis1 = [0, ]
    if Enquiry(lis1):
        print("Not Empty")
    else:
        print("Empty")

    Output:

    Empty

    Making the Numpythonic way work

    1. If we have a numpy array then correct method in all cases, is to use if .size. This size checks the size of the arrays and return True or False accordingly.
      Example:




      # Numpythonic way to check emptiness
      # Use of size
      import numpy
      def Enquiry(lis1):
          return(numpy.array(lis1))
            
      # Driver Code
      lis1 = []
      if Enquiry(lis1).size:
          print("Not Empty")
      else:
          print("Empty")

      Output:

      Empty
    2. This example shows the other case with a single 0 element, which failed in the previous cases.




      # Numpythonic way to check emptiness
      # Use of size
      import numpy
      def Enquiry(lis1):
          return(numpy.array(lis1))
            
      # Driver Code
      lis1 = [0, ]
      if Enquiry(lis1).size:
          print("Not Empty")
      else:
          print("Empty")

      Output:

      Not Empty

    For more reference visit PEP8 style guide.

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