[英]Counting Inversion for a large text file
该代码的目的是找出总数。 数组的反转。 我的代码成功运行。 成功测试了6个元素(所有元素以逆序从最高位开始),反转计数=15。此外,成功测试了10个元素(所有元素以逆序从高位开始),反转计数= 45但是,对于大型包含100k整数的文件,耗时将近25秒。 这是预期的吗? 请建议还是我进一步缩短执行时间? 我只是对常规的合并排序算法进行了一些细微的调整(即用于计算反转总数的行),如何进一步减少总体运行时间?
def mergeSort(final_list):
global total_count
if len(final_list)>1:
mid_no=len(final_list)//2
left_half=final_list[:mid_no]
right_half=final_list[mid_no:]
mergeSort(left_half)
mergeSort(right_half)
'''Below code is for merging the lists'''
i=j=k=0 #i is index for left half, j for the right half and k for the resultant list
while i<len(left_half) and j<len(right_half):
if left_half[i] < right_half[j]:
final_list[k]=left_half[i]
i+=1
k+=1
else:
final_list[k]=right_half[j]
print 'total count is'
print total_count
#total_count+=len(left_half)-i
total_count+=len(left_half[i:])
print 'total_count is '
print total_count
print 'pairs are '
print str(left_half[i:])+' with '+str(right_half[j])
j+=1
k+=1
while i<len(left_half):
final_list[k]=left_half[i]
k+=1
i+=1
while j<len(right_half):
final_list[k]=right_half[j]
j+=1
k+=1
'''Code for list merge ends'''
#temp_list=[45,21,23,4,65]
#temp_list=[1,5,2,3,4,6]
#temp_list=[6,5,4,3,2,1]
#temp_list=[1,2,3,4,5,6]
#temp_list=[10,9,8,7,6,5,4,3,2,1]
#temp_list=[1,22,3,4,66,7]
temp_list=[]
f=open('temp_list.txt','r')
for line in f:
temp_list.append(int(line.strip()))
print 'list is '
print temp_list
print 'list ends'
print temp_list[0]
print temp_list[-1]
'''import time
time.sleep(1000)
print 'hhhhhhhhhh'
'''
total_count=0
mergeSort(temp_list)
print temp_list
我找到了(并已通过个人资料验证)
#total_count+=len(left_half[i:])
total_count += len(left_half) - i
left_half [i:]在递归函数的主循环中多次创建一个包含多个元素的副本的新列表。 这是对拼接的巧妙运用,但副作用却在损害您的性能。
这是我分解功能以对其进行分析的方式:
def so_merge (final_list, left_half, right_half):
global total_count
i=j=k=0 #i is index for left half, j for the right half and k for the resultant list
while i<len(left_half) and j<len(right_half):
if left_half[i] < right_half[j]:
final_list[k]=left_half[i]
i+=1
k+=1
else:
final_list[k]=right_half[j]
count1 = get_incriment_bad(left_half, i)
count2 = get_incriment_good(left_half, i)
if count1 != count2:
raise ValueError
total_count += count1
j+=1
k+=1
finish_left(final_list, left_half, i, k)
finish_right(final_list, right_half, j, k)
结果表明,花len(left_half [i:])花了19.574秒
ncalls tottime percall cumtime percall filename:lineno(function)
199999/1 0.805 0.000 29.562 29.562 week1.py:124(so_mergesort)
99999 7.496 0.000 28.735 0.000 week1.py:104(so_merge)
776644 19.512 0.000 19.574 0.000 week1.py:101(get_incriment_bad)
776644 0.839 0.000 0.895 0.000 week1.py:98(get_incriment_good)
5403164 0.382 0.000 0.382 0.000 {len}
99999 0.273 0.000 0.286 0.000 week1.py:92(finish_right)
99999 0.255 0.000 0.266 0.000 week1.py:86(finish_left)
1 0.000 0.000 0.000 0.000 {method 'disable' of '_lsprof.Profiler' objects}
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