如您所见，以下代码的结果是不公平的，因为它不可能是迭代器与“ return”或“ if y：”花费相同的时间。为什么？我该如何解决？
在某些测试中，我的代码太重了，因此我尝试检查每一行的“内存支出”。
def hmd(num):
if num==1:
return 1
def resheto(num):
t = list(range(num+1))
t[1]=0
for i in t:
if t[i]:
for u in range(i**2, num+1, i):
t[u]=0
end = [i for i in range(2,num+1) if t[i]]
end = list(filter(lambda x: num%x==0, end))
return end
l = resheto(num)
def end(num, l, minw, u=[]):
u.append(minw)
num=num//minw
if num==1:
return u
ent = list(filter(lambda x: num%x==0, l))
minw=ent[0]
return end(num, l, minw, u)
o = end(num, l, l[0])
def hi(o, j=[]):
if len(o)==0:
return j
y = len(list(filter(lambda x: o[0]==x, o)))
u = list(filter(lambda x: o[0]!=x, o))
if y:
j.append('{p}^{y}'.format(p=o[0], y=y))
else:
j.append(str(o[0]))
return hi(u, j)
return ('*'.join(hi(o))).replace('^1','')
from dgdgd_zooz_memory import hmd
%mprun -f hmd hmd(100)
---结果---
Filename: /Users/Pro/Desktop/python/sirius/dgdgd_zooz_memory.py
1 56.5 MiB 56.5 MiB def hmd(num):
2 56.5 MiB 0.0 MiB if num==1:
3 return 1
4 56.5 MiB 0.0 MiB def resheto(num):
5 56.5 MiB 0.0 MiB t = list(range(num+1))
6 56.5 MiB 0.0 MiB t[1]=0
7 56.5 MiB 0.0 MiB for i in t:
8 56.5 MiB 0.0 MiB if t[i]:
9 56.5 MiB 0.0 MiB for u in range(i**2, num+1, i):
10 56.5 MiB 0.0 MiB t[u]=0
11 56.5 MiB 0.0 MiB end = [i for i in range(2,num+1) if t[i]]
12 56.5 MiB 0.0 MiB end = list(filter(lambda x: num%x==0, end))
13 56.5 MiB 0.0 MiB return end
14 56.5 MiB 0.0 MiB l = resheto(num)
15 56.5 MiB 0.0 MiB def end(num, l, minw, u=[]):
16 56.5 MiB 0.0 MiB u.append(minw)
17 56.5 MiB 0.0 MiB num=num//minw
18 56.5 MiB 0.0 MiB if num==1:
19 56.5 MiB 0.0 MiB return u
20 56.5 MiB 0.0 MiB ent = list(filter(lambda x: num%x==0, l))
21 56.5 MiB 0.0 MiB minw=ent[0]
22 56.5 MiB 0.0 MiB return end(num, l, minw, u)
23 56.5 MiB 0.0 MiB o = end(num, l, l[0])
24 56.5 MiB 0.0 MiB def hi(o, j=[]):
25 56.5 MiB 0.0 MiB if len(o)==0:
26 56.5 MiB 0.0 MiB return j
27 56.5 MiB 0.0 MiB y = len(list(filter(lambda x: o[0]==x, o)))
28 56.5 MiB 0.0 MiB u = list(filter(lambda x: o[0]!=x, o))
29 56.5 MiB 0.0 MiB if y:
30 56.5 MiB 0.0 MiB j.append('{p}^{y}'.format(p=o[0], y=y))
31 else:
32 j.append(str(o[0]))
33 56.5 MiB 0.0 MiB return hi(u, j)
34 56.5 MiB 0.0 MiB return ('*'.join(hi(o))).replace('^1','')