Why is "try ... except KeyError ..." slower than "dict.get"?

Question

Long time before, someone told me

d = {}
try:
    a = d['a']
    # do something
except KeyError:
    # do something else

is faster than

a = d.get('a')
if a:
    # do something
else:
    # do something else

But today, after I did a simple test, I found the result just on the contrary:

import time

d = {}
n = 0
t1 = time.time()
for i in range(1000000):
    try:
        a = d['a']
    except KeyError:
        n += 1
print(time.time() - t1)
>>> 0.4676947593688965

import time

d = {}
n = 0
t1 = time.time()
for i in range(1000000):
    a = d.get('a')
    if a is None:
        n += 1
print(time.time() - t1)
>>> 0.3045947551727295

So why does it happened? In my imagine, d.get should also do the similar logic that judge whether the key is existing in the dict.

Answer 1

The rule you were taught applies when the exceptional case is the uncommon case. If you're usually going to find the value, d['a'] wrapped in a try / except KeyError: that rarely invokes the exception handling machinery will win; if you're usually not going to find the value, the overhead of throwing and catching the exception will outweigh the higher relative overhead of using generic method dispatch ( d.get('a') ) vs. the more direct syntax-supported approach ( d['a'] ). That's especially true in modern (3.7+) CPython, where method calls got some additional optimization , reducing the overhead of d.get('a') .

As you note in the comments , in cases where the lookup succeeds , get remains the same speed, while d['a'] with an unused except KeyError: speeds up to be significantly faster. It's also more correct in many cases; if the dict had keys that mapped to None , your get -based code would wouldn't distinguish "key not found" from "key maps to None ", which might be what you want, but often isn't.