Should I store a calculation in a variable if it will be used a lot?

Question

If I have a function to, for example, check if list1 is a sublist of list2, which option is better:

Option 1:

Answer 1

Storing local is better because the lookup that python does is faster. It even pays off to store functions locally.

Performance questions are best answered by timing them - you can measure this using timeit:

import timeit

def noTempFunc():
    for _ in range(200):
        max([1,4,5,6])

def tempFunc():
    m = max
    for _ in range(200):
        m([1,4,5,6])


print(timeit.timeit(noTempFunc, number=1000))   #  0.055301458000030834
print(timeit.timeit(tempFunc, number=1000))     #  0.049811941999905684 : 11% faster

In this case the max() of the global context only has to be looked up once, and further lookups are done locally which is ~ 11% faster based on these numbers.

It pays up if you use your local m() multiple times.

---

In your case it would be sensible to cache the len_list1 = len(list1) - as it is used lots of time and does not change.

To make it more readable, you can consider:

def isSublist(list1, list2):
    """Checks if list2 is a sublist of list1"""
    len_part = len(list2)  # reused inside the list comp, only "calulated" once
    return any( x == list2 for x in (list1[i:i+len_part] 
                                     for i in range(len(list1)-len_part+1) ))


print(isSublist([1,2,3,4],[1]))
print(isSublist([1,2,3,4],[2,3]))
print(isSublist([1,2,3,4],[1,2,4]))
print(isSublist([1,2,3,4],[1,2,3,4]))

Output:

True
True
False
True

Lookups:

• any() && How exactly does the python any() function work?
• Split a python list into other "sublists" i.e smaller lists

---

Your faster version with cached length as well (based on Scott Mermelstein answer):

def isSublist1a(list1,list2): # cached length as well
    l1 = len(list1)
    for i in range(len(list2)):
        part=list2[i:]
        if len(part)<l1:
            return False
        if list1==part[:l1]:
            return True
    return False

list1=list(range(1000))
list2=list(range(400,420))
import timeit
print(timeit.timeit('isSublist1(list2,list1)', globals=globals(),number=1000))
print(timeit.timeit('isSublist1a(list2,list1)', globals=globals(),number=1000))
print(timeit.timeit('isSublist2(list2,list1)', globals=globals(),number=1000))

delivers (2 executions):

0.08652938600062043  # cached part
0.08017484299944044  # cached part + cached list1 lenght - slightly faster
0.15090413599955355  # non-cached version

0.8882850420004615   # cached part
0.8294611960000111   # cached part + cached list1 lenght - slightly faster
1.5524438030006422   # non-cached version

Answer 2

Supplementary to Patrick's excellent answer, let's try timeit with your actual code:

>>> def isSublist1(list1,list2):                                                                                                                                                                                                                              
...   for i in range(len(list2)):                                                                                                                                                                                                                             
...     part=list2[i:]                                                                                                                                                                                                                                        
...     if len(part)<len(list1):                                                                                                                                                                                                                              
...       return False                                                                                                                                                                                                                                        
...     if list1==part[:len(list1)]:
...       return True                                                                                                                                                                                                                                         
...   return False                                                                                                                                                                                                                                            
... 
>>> def isSublist2(list1,list2):
...   for i in range(len(list2)):
...     if len(list2[i:])<len(list1):
...       return False
...     if list1==list2[i:][:len(list1)]:
...       return True
...   return False
... 
>>> list1=list(range(10000))
>>> list2=list(range(4000,4020))
>>> import timeit
>>> timeit.timeit('isSublist1(list2,list1)', globals=globals(),number=100)
6.420147094002459
>>> timeit.timeit('isSublist2(list2,list1)', globals=globals(),number=100)
12.455138996010646

So on my system, the amount of time needed with your temporary variable is just about half the time needed without it.

I don't know the nature of your lists and sublists; you may want to change what list1 and list2 are to be a good reflection of how your code will be used, but at least on my side, it looks like saving that temporary variable is a very good idea.

Incidentally, let's do another interesting experiment:

>>> def isSublist3(list1,list2):
...   ln = len(list1)
...   for i in range(len(list2)):
...     part=list2[i:]
...     if len(part)<ln:
...       return False
...     if list1==part[:ln]:
...       return True
...   return False
... 
>>> timeit.timeit('isSublist1(list2,list1)',globals=globals(),number=100); timeit.timeit('isSublist3(list2,list1)',globals=globals(),number=100)
6.549526696035173
6.481004184985068

I ran it a few more times to see what I'd get:

6.470875242026523 6.463623657007702

6.151073662971612 5.787795798969455

5.685607994964812 5.655005165026523

6.439315696014091 6.372227535001002

Note that each time, the cached length takes less time than the non-cached length, though you don't get nearly the performance improvement you had by caching the slicing.

Note also that it's important not to draw too many conclusions from a single run of timeit. There are a lot of other variables affecting the timing, (in my case, pretty clearly, something happened to make it drop from 6.4 to 5.7 - in the middle of a single test!) so if you want to come up with a good rule you can count on, test several times to make sure you get consistent results.