if not a:
print("List is empty")
Using the implicit booleanness of the empty list is quite Pythonic.
if not a:
print("List is empty")
Using the implicit booleanness of the empty list is quite Pythonic.
The Pythonic way to do it is from the PEP 8 style guide.
For sequences, (strings, lists, tuples), use the fact that empty sequences are false:
# Correct: if not seq: if seq: # Wrong: if len(seq): if not len(seq):
python - How to check if array is not empty? - Stack Overflow
python - Check if list or array element is empty - Stack Overflow
How do I check if this list is "empty"?
python - How can I check whether a numpy array is empty or not? - Stack Overflow
Is an array of NaN values empty?
Why can I not use if array?
How do I check an empty dimension?
There's no mention of numpy in the question. If by array you mean list, then if you treat a list as a boolean it will yield True if it has items and False if it's empty.
l = []
if l:
print "list has items"
if not l:
print "list is empty"
with a as a numpy array, use:
if a.size:
print('array is not empty')
(in Python, objects like [1,2,3] are called lists, not arrays.)
I'm completely stumped on a computer science assignment because our lecturer has given us zero information. We're supposed to have the user input a list and check if the list is in decreasing order and print a suitable message if it is. That part works fine. However, if the list is "empty" (I'll explain the quotations in a second) then it should print the list is empty.
The problem is that, according to our brief we're supposed to use .split(",") in order to allow the user to separate their numbers with commas. The problem lies in the fact that even if a user hits enter without inputting anything, the list is still '1' long with just an empty string. I've been scouring the web to find out how to check if a list is empty. But all the tutorials have so far just been in relation to actually empty lists (as in, list = []). One did use the all() function to check that all items in the list were strings, which did work and let me print that the list is empty, but broke the original function. Could someone please help me I'm tearing my hair out with this.
--
def checkOrder():
# checks, etc.
integer_list = []
listLength = len(integer_list)
increasing = 0
decreasing = 0
singleValue = 0
# Converting inputted numbers into a split list
userInput = input("Please enter your numbers seperated by a comma (,):")
inputtedStrings = userInput.split(",")
for number in inputtedStrings:
inputtedIntegers = int(number)
integer_list.append(inputtedIntegers)
# Enumerating list
for i, val in enumerate(integer_list[:-1]):
if val <= integer_list[i+1]:
increasing = 1
elif val >= integer_list[i+1]:
decreasing = 1
# Check for a single value
if len(integer_list) == 1:
singleValue = 1
if len(integer_list) == 1 and
print("The list is empty.")
if increasing == 1:
print("The numbers are in increasing order.")
elif decreasing == 1 or singleValue == 1:
print("The numbers are in decreasing order.")
checkOrder()
--
You can always take a look at the .size attribute. It is defined as an integer, and is zero (0) when there are no elements in the array:
import numpy as np
a = np.array([])
if a.size == 0:
# Do something when `a` is empty
One caveat, though.
Note that np.array(None).size returns 1!
This is because a.size is equivalent to np.prod(a.shape),
np.array(None).shape is (), and an empty product is 1.
>>> import numpy as np
>>> np.array(None).size
1
>>> np.array(None).shape
()
>>> np.prod(())
1.0
Therefore, I use the following to test if a NumPy array has elements:
>>> def elements(array):
... return array.ndim and array.size
>>> elements(np.array(None))
0
>>> elements(np.array([]))
0
>>> elements(np.zeros((2,3,4)))
24