The efficiency of Python keyword “in” may depend on the data structure it is applied to. In this post I will show you the efficiency difference of “in” when used with lists and dictionaries (hash tables).
First, I composed the following two python scripts to test the elapsed time when using lists and dictionaries.
The script for list creates a list of range(10000) and check for 0-9999 whether they are in the list using keyword “in”.
import time
in_list = range(10000)
counter = 0
start = time.clock()
for i in range(10000):
if i in in_list:
counter += 1
end = time.clock()
print "hit number:"
print counter
print "elapsed time:"
print (end-start)
The script for dictionary first creates a dictionary consisting 10000 key:value pairs. Then it tests the time needed to check whether 0-9999 are in the dictionary.
import time
in_dict = {k:v for k,v in zip(range(10000),range(10000))}
counter = 0
start = time.clock()
for i in range(10000):
if i in in_dict:
counter += 1
end = time.clock()
print "hit number:"
print counter
print "elapsed time:"
print (end-start)
The results are as follow, respectively.
list:
hit number:
10000
elapsed time:
0.694395
dictionary:
hit number:
10000
elapsed time:
0.001719
The results show that although the keyword is “in” for both of the data structures, the efficiency varies. In another word, the efficiency of Python keyword “in” depends on the data structure it is applied to.
peidongwang 