TallPaul - 1 month ago 5x

Python Question

I recently posted a question using a lambda function and in a reply someone had mentioned lambda is going out of favor, to use list comprehensions instead. I am relatively new to Python. I ran a simple test:

`import time`

S=[x for x in range(1000000)]

T=[y**2 for y in range(300)]

#

#

time1 = time.time()

N=[x for x in S for y in T if x==y]

time2 = time.time()

print 'time diff [x for x in S for y in T if x==y]=', time2-time1

#print N

#

#

time1 = time.time()

N=filter(lambda x:x in S,T)

time2 = time.time()

print 'time diff filter(lambda x:x in S,T)=', time2-time1

#print N

#

#

#http://snipt.net/voyeg3r/python-intersect-lists/

time1 = time.time()

N = [val for val in S if val in T]

time2 = time.time()

print 'time diff [val for val in S if val in T]=', time2-time1

#print N

#

#

time1 = time.time()

N= list(set(S) & set(T))

time2 = time.time()

print 'time diff list(set(S) & set(T))=', time2-time1

#print N #the results will be unordered as compared to the other ways!!!

#

#

time1 = time.time()

N=[]

for x in S:

for y in T:

if x==y:

N.append(x)

time2 = time.time()

print 'time diff using traditional for loop', time2-time1

#print N

They all print the same N so I commented that print stmt out (except the last way it's unordered), but the resulting time differences were interesting over repeated tests as seen in this one example:

`time diff [x for x in S for y in T if x==y]= 54.875`

time diff filter(lambda x:x in S,T)= 0.391000032425

time diff [val for val in S if val in T]= 12.6089999676

time diff list(set(S) & set(T))= 0.125

time diff using traditional for loop 54.7970001698

So while I find list comprehensions on the whole easier to read, there seems to be some performance issues at least in this example.

So, two questions:

- Why is lambda etc being pushed aside?
- For the list comprehension ways, is there a more efficient implementation and how would you KNOW it's more efficient without testing? I mean, lambda/map/filter was supposed to be less efficient because of the extra function calls, but it seems to be MORE efficient.

Paul

Answer

Your tests are doing very different things. With S being 1M elements and T being 300:

```
[x for x in S for y in T if x==y]= 54.875
```

This option does 300M equality comparisons.

```
filter(lambda x:x in S,T)= 0.391000032425
```

This option does 300 linear searches through S.

```
[val for val in S if val in T]= 12.6089999676
```

This option does 1M linear searches through T.

```
list(set(S) & set(T))= 0.125
```

This option does two set constructions and one set intersection.

The differences in performance between these options is much more related to the algorithms each one is using, *rather* than any difference between list comprehensions and `lambda`

.

Source (Stackoverflow)

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