Python None comparison: should I use "is" or ==?
PythonComparisonNonetypePython Problem Overview
My editor warns me when I compare my_var == None
, but no warning when I use my_var is None
.
I did a test in the Python shell and determined both are valid syntax, but my editor seems to be saying that my_var is None
is preferred.
Is this the case, and if so, why?
Python Solutions
Solution 1 - Python
Summary:
Use is
when you want to check against an object's identity (e.g. checking to see if var
is None
). Use ==
when you want to check equality (e.g. Is var
equal to 3
?).
Explanation:
You can have custom classes where my_var == None
will return True
e.g:
class Negator(object):
def __eq__(self,other):
return not other
thing = Negator()
print thing == None #True
print thing is None #False
is
checks for object identity. There is only 1 object None
, so when you do my_var is None
, you're checking whether they actually are the same object (not just equivalent objects)
In other words, ==
is a check for equivalence (which is defined from object to object) whereas is
checks for object identity:
lst = [1,2,3]
lst == lst[:] # This is True since the lists are "equivalent"
lst is lst[:] # This is False since they're actually different objects
Solution 2 - Python
is
is generally preferred when comparing arbitrary objects to singletons like None
because it is faster and more predictable. is
always compares by object identity, whereas what ==
will do depends on the exact type of the operands and even on their ordering.
This recommendation is supported by PEP 8, which explicitly states that "comparisons to singletons like None should always be done with is
or is not
, never the equality operators."
Solution 3 - Python
PEP 8 defines that it is better to use the is
operator when comparing singletons.
Solution 4 - Python
I recently encountered where this can go wrong.
import numpy as np
nparray = np.arange(4)
# Works
def foo_is(x=None):
if x is not None:
print(x[1])
foo_is()
foo_is(nparray)
# Code below raises
# ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
def foo_eq(x=None):
if x != None:
print(x[1])
foo_eq()
foo_eq(nparray)
I created a function that optionally takes a numpy array as argument and changes if it is included. If I test for its inclusion using inequality operators !=
, this raises a ValueError (see code above). If I use is not none
, the code works correctly.
Solution 5 - Python
Another instance where "==" differs from "is". When you pull information from a database and check if a value exists, the result will be either a value or None.
Look at the if and else below. Only "is" works when the database returns "None". If you put == instead, the if statement won't work, it will go straight to else, even though the result is "None". Hopefully, I am making myself clear.
conn = sqlite3.connect('test.db')
c = conn.cursor()
row = itemID_box.get()
# pull data to be logged so that the deletion is recorded
query = "SELECT itemID, item, description FROM items WHERE itemID LIKE '%" + row + "%'"
c.execute(query)
result = c.fetchone()
if result is None:
# log the deletion in the app.log file
logging = logger('Error')
logging.info(f'The deletion of {row} failed.')
messagebox.showwarning("Warning", "The record number is invalid")
else:
# execute the deletion
c.execute("DELETE from items WHERE itemID = " + row)
itemID_box.delete(0, tk.END)
messagebox.showinfo("Warning", "The record has been deleted")
conn.commit()
conn.close()
Solution 6 - Python
We can take the list for example. Look here:
a = list('Hello')
b = a
c = a[:]
All have the same value:
['H', 'e', 'l', 'l', 'o']
['H', 'e', 'l', 'l', 'o']
['H', 'e', 'l', 'l', 'o']
is
refers to the actual memory slot, whether they are the specific object of interest:
print(a is b)
print(a is c)
Now we get the desired result.
True
False
PEP 8 also mentions this, saying "comparisons to singletons like None should always be done with is or is not, never the equality operators."
is
is also faster.