python 函数的分析

Question

您知道如何使该功能更加省时吗？

def c(n):
    word = 32
    #l = []
    c = 0
    for i in range(0, 2**word):
        #print(str(bin(i)))#.count('1')
        if str(bin(i)).count('1') == n:
            c = c + 1 
            print(c)

        if i == 2**28:
            print('6 %')
        if i == 2**29:
            print('12 %')
        if i == 2**30:
            print('25 %')
        if i == 2**31:
            print('50 %')
        if i == 2**32:
            print('100 %')
    return c



135274023 function calls in 742.161 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1  391.662  391.662  742.161  742.161 <pyshell#3>:1(c)
        1    0.000    0.000  742.161  742.161 <string>:1(<module>)
     4816    0.014    0.000    0.014    0.000 rpc.py:149(debug)
      688    0.010    0.000    3.162    0.005 rpc.py:208(remotecall)
      688    0.017    0.000    0.107    0.000 rpc.py:218(asynccall)
      688    0.019    0.000    3.043    0.004 rpc.py:238(asyncreturn)
      688    0.002    0.000    0.002    0.000 rpc.py:244(decoderesponse)
      688    0.007    0.000    3.018    0.004 rpc.py:279(getresponse)
      688    0.006    0.000    0.010    0.000 rpc.py:287(_proxify)
      688    0.025    0.000    3.000    0.004 rpc.py:295(_getresponse)
      688    0.002    0.000    0.002    0.000 rpc.py:317(newseq)
      688    0.023    0.000    0.062    0.000 rpc.py:321(putmessage)
      688    0.007    0.000    0.011    0.000 rpc.py:546(__getattr__)
      688    0.002    0.000    0.002    0.000 rpc.py:587(__init__)
      688    0.004    0.000    3.166    0.005 rpc.py:592(__call__)
     1376    0.008    0.000    0.011    0.000 threading.py:1012(current_thread)
      688    0.004    0.000    0.019    0.000 threading.py:172(Condition)
      688    0.009    0.000    0.015    0.000 threading.py:177(__init__)
      688    0.019    0.000    2.962    0.004 threading.py:226(wait)
      688    0.002    0.000    0.002    0.000 threading.py:45(__init__)
      688    0.002    0.000    0.002    0.000 threading.py:50(_note)
      688    0.004    0.000    0.004    0.000 threading.py:88(RLock)
      688    0.004    0.000    0.004    0.000 {built-in method allocate_lock}
 67620326  162.442    0.000  162.442    0.000 {built-in method bin}
      688    0.007    0.000    0.007    0.000 {built-in method dumps}
        1    0.000    0.000  742.161  742.161 {built-in method exec}
     1376    0.003    0.000    0.003    0.000 {built-in method get_ident}
     1376    0.004    0.000    0.004    0.000 {built-in method isinstance}
     2064    0.005    0.000    0.005    0.000 {built-in method len}
      688    0.002    0.000    0.002    0.000 {built-in method pack}
      344    0.009    0.000    3.187    0.009 {built-in method print}
      688    0.008    0.000    0.008    0.000 {built-in method select}
      688    0.003    0.000    0.003    0.000 {method '_acquire_restore' of '_thread.RLock' objects}
      688    0.002    0.000    0.002    0.000 {method '_is_owned' of '_thread.RLock' objects}
      688    0.002    0.000    0.002    0.000 {method '_release_save' of '_thread.RLock' objects}
      688    0.003    0.000    0.003    0.000 {method 'acquire' of '_thread.RLock' objects}
     1376    2.929    0.002    2.929    0.002 {method 'acquire' of '_thread.lock' objects}
      688    0.002    0.000    0.002    0.000 {method 'append' of 'list' objects}
 67620325  184.869    0.000  184.869    0.000 {method 'count' of 'str' objects}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
      688    0.002    0.000    0.002    0.000 {method 'get' of 'dict' objects}
      688    0.002    0.000    0.002    0.000 {method 'release' of '_thread.RLock' objects}
      688    0.015    0.000    0.015    0.000 {method 'send' of '_socket.socket' objects}

我试图实现的是计算从 0 到 2**32 有多少个数字在其二进制表示中具有 n 个 1 。

Answer 1

您正在计算有多少 32 位数字具有给定数量的 1。这个数字就是二项式系数 32个选择位，并且可以计算和：

from math import factorial
print factorial(32) // (factorial(bits) * factorial(32-bits))