Question
Suppose we would like to create an array with 100 elements.
2 types of elements will be tested
Noneintegersfrom 0 to 99
Methods
3 ways to initialize an array
- By append
def byAppend():
a = []
for i in xrange(100):
a.append(None)
def byAppendInit():
a = []
for i in xrange(100):
a.append(i)
- By list comprehension
def byListComprehension():
[None for i in xrange(100)]
def byComprehensionInit():
[i for i in xrange(100)]
- By multiplication
def byMultiplication():
[None] * 100
def byMultiplicationInit():
a = [None] * 100
for i in xrange(100):
a[i] = i
Results
import timeit
print 'By append'
print(timeit.timeit('byAppend()', setup="from __main__ import byAppend"))
print 'By append init'
print(timeit.timeit('byAppendInit()', setup="from __main__ import byAppendInit"))
print 'By list comprehension'
print(timeit.timeit('byListComprehension()', setup="from __main__ import byListComprehension"))
print 'By list comprehensionInit'
print(timeit.timeit('byComprehensionInit()', setup="from __main__ import byComprehensionInit"))
print 'By byMultiplication'
print(timeit.timeit('byMultiplication()', setup="from __main__ import byMultiplication"))
print 'By byMultiplication init'
print(timeit.timeit('byMultiplicationInit()', setup="from __main__ import byMultiplicationInit"))
Output from consoles
C:\Python27\python.exe C:/Users/Mond/PycharmProjects/hackercup/round1/test.py
By append
10.2452175728
By append init
9.83314044088
By list comprehension
4.45988583823
By list comprehensionInit
4.37547215318
By byMultiplication
0.840903578289
By byMultiplication init
5.83873665989
Conclusion
- If you need to create a “fixed sized” array only, multiplication is the fastest.
- If you need to initialize array with some constants, list comprehension performs better.
- Never ever using the way for appending.
Gotcha
For 2d array initialization, you must be careful not to do it via multiplication.
Below are codes for initializing a 2D array with 3 rows and try to add a string
hello to the first array
a = [[]] * 3
a[0].append('hello')
print a
What
a will be output is[['hello'], ['hello'], ['hello']]
instead of
[['hello'], [], []]
Therefore, we can see initializing 2D array via multiplication is not a good way as they are all the same array even though arrays seems to be multiplied (initialized).
The proper way is using
list comprehension# The underscore `_` means I do not care what value it is
a = [[] for _ in xrange(3)]
a[0].append('hello')
print a
Then,
a will be output as[['hello'], [], []]
