Lastor already said what I was going to point out so I am not going to repeat that. I'll just add some other things.
I tried timing your solution with a bunch of other solutions I came up with. Among these the one with best time-memory combination should be the function mysort3 as it gave me best timing in nearly all cases. I am still looking about proper timing proper timing in Python. You can try putting in different test cases in the function tests to test the timing for yourself.
def mysort(words):
mylist1 = sorted([i for i in words if i[:1] == "s"])
mylist2 = sorted([i for i in words if i[:1] != "s"])
list = mylist1 + mylist2
return list
def mysort3(words):
ans = []
p = ans.append
q = words.remove
words.sort()
for i in words[:]:
if i[0] == 's':
p(i)
q(i)
return ans + words
def mysort4(words):
ans1 = []
ans2 = []
p = ans1.append
q = ans2.append
for i in words:
if i[0] == 's':
p(i)
else:
q(i)
ans1.sort()
ans2.sort()
return ans1 + ans2
def mysort6(words):
return ( sorted([i for i in words if i[:1] == "s"]) +
sorted([i for i in words if i[:1] != "s"])
)
if __name__ == "__main__":
from timeit import Timer
def test(f):
f(['a','b','c','abcd','s','se', 'ee', 'as'])
print Timer(lambda: test(mysort)).timeit(number = 10000)
print Timer(lambda: test(mysort3)).timeit(number = 10000)
print Timer(lambda: test(mysort4)).timeit(number = 10000)
print Timer(lambda: test(mysort6)).timeit(number = 10000)
Lastor already said what I was going to point out so I am not going to repeat that. I'll just add some other things.
I tried timing your solution with a bunch of other solutions I came up with. Among these the one with best time-memory combination should be the function mysort3 as it gave me best timing in nearly all cases. I am still looking about proper timing in Python. You can try putting in different test cases in the function tests to test the timing for yourself.
def mysort(words):
mylist1 = sorted([i for i in words if i[:1] == "s"])
mylist2 = sorted([i for i in words if i[:1] != "s"])
list = mylist1 + mylist2
return list
def mysort3(words):
ans = []
p = ans.append
q = words.remove
words.sort()
for i in words[:]:
if i[0] == 's':
p(i)
q(i)
return ans + words
def mysort4(words):
ans1 = []
ans2 = []
p = ans1.append
q = ans2.append
for i in words:
if i[0] == 's':
p(i)
else:
q(i)
ans1.sort()
ans2.sort()
return ans1 + ans2
def mysort6(words):
return ( sorted([i for i in words if i[:1] == "s"]) +
sorted([i for i in words if i[:1] != "s"])
)
if __name__ == "__main__":
from timeit import Timer
def test(f):
f(['a','b','c','abcd','s','se', 'ee', 'as'])
print Timer(lambda: test(mysort)).timeit(number = 10000)
print Timer(lambda: test(mysort3)).timeit(number = 10000)
print Timer(lambda: test(mysort4)).timeit(number = 10000)
print Timer(lambda: test(mysort6)).timeit(number = 10000)
Lastor already said what I was going to point out so I am not going to repeat that. I'll just add some other things.
I tried timing your solution with a bunch of other solutions I came up with. Among these the one with best time-memory combination should be the function mysort3 as it gave me best timing in nearly all cases. I am still looking about proper timing in Python. You can try putting in different test cases in the function tests to test the timing for yourself.
def mysort(words):
mylist1 = sorted([i for i in words if i[:1] == "s"])
mylist2 = sorted([i for i in words if i[:1] != "s"])
list = mylist1 + mylist2
return list
def mysort3(words):
ans = []
p = ans.append
q = words.remove
words.sort()
for i in words[:]:
if i[0] == 's':
p(i)
q(i)
return ans + words
def mysort4(words):
ans1 = []
ans2 = []
p = ans1.append
q = ans2.append
for i in words:
if i[0] == 's':
p(i)
else:
q(i)
ans1.sort()
ans2.sort()
return ans1 + ans2
def mysort6(words):
return ( sorted([i for i in words if i[:1] == "s"]) +
sorted([i for i in words if i[:1] != "s"])
)
if __name__ == "__main__":
from timeit import Timer
def test(f):
f(['a','b','c','abcd','s','se', 'ee', 'as'])
print Timer(lambda: test(mysort)).timeit(number = 10000)
print Timer(lambda: test(mysort3)).timeit(number = 10000)
print Timer(lambda: test(mysort4)).timeit(number = 10000)
print Timer(lambda: test(mysort6)).timeit(number = 10000)
Lastor already said what I was going to point out so I am not going to repeat that. I'll just add some other things.
I tried timing your solution with a bunch of other solutions I came up with. Among these the one with best time-memory combination should be the function mysort3 as it gave me best timing in nearly all cases. I am still looking about proper timing in Python. You can try putting in different test cases in the function tests to test the timing for yourself.
def mysort(words):
mylist1 = sorted([i for i in words if i[:1] == "s"])
mylist2 = sorted([i for i in words if i[:1] != "s"])
list = mylist1 + mylist2
return list
def mysort3(words):
ans = []
p = ans.append
q = words.remove
words.sort()
for i in words[:]:
if i[0] == 's':
p(i)
q(i)
return ans + words
def mysort4(words):
ans1 = []
ans2 = []
p = ans1.append
q = ans2.append
for i in words:
if i[0] == 's':
p(i)
else:
q(i)
ans1.sort()
ans2.sort()
return ans1 + ans2
def mysort6(words):
return ( sorted([i for i in words if i[:1] == "s"]) +
sorted([i for i in words if i[:1] != "s"])
)
if __name__ == "__main__":
from timeit import Timer
def test(f):
f(['a','b','c','abcd','s','se', 'ee', 'as'])
print Timer(lambda: test(mysort)).timeit(number = 10000)
print Timer(lambda: test(mysort3)).timeit(number = 10000)
print Timer(lambda: test(mysort4)).timeit(number = 10000)
print Timer(lambda: test(mysort6)).timeit(number = 10000)