如何获得一个重复n次的数字?

问题描述

我有一个数字很多的文件

0.98
0.23
0.10
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
10.3
11.9
0.56
...

我要打印连续10次(最少)重复数字0的行数。考虑以上输入,输出将是:4(对于行号4,因为0连续重复10次)。文件list.txt是一个巨大的文件。我是Python新手。我该如何删除以下脚本中的错误

import ast
values = open("list.txt","r")
values = list(map(int,ast.literal_eval(values.read().strip())))
count=0
length=""
if len(values)>1:
    for i in range(1,len(values)):
       if values[i-1]==values[i]:
          count+=1
       else :
           length += values[i-1]+" repeats "+str(count)+","
           count=1
    length += ("and "+values[i]+" repeats "+str(count))
else:
    i=0
    length += ("and "+values[i]+" repeats "+str(count))
print (length)

解决方法

逐行读取和评估文件。如果找到了模式,则循环中断,停止读取文件

import ast
count = 0
lineNb = -1
found = False # False by default
with open("list.txt") as f:
    for i,line in enumerate(f): # loop over lines,one-by-one
        value = ast.literal_eval(line)
        if value == 0:
            if count == 0: # first occurrence
                lineNb = i # set potential lineNb
            count += 1     # increment counter
            if count == 10: # desired condition
                found = True # now we know we have found the pattern
                break        # break the for loop
        else: # not 0
            count = 0 # reset counter

print(found,lineNb) # (True,3) # lineNb is zero-based,3 = 4th line
,
with open('consecutive.txt') as f:
    c = 0
    for i,line in enumerate(f):
        if float(line)==0.0:
            c+=1
            if c == 10:
                print(i-8)
                break
        else:
            c=0

输出

4