如何得到一个重复n次的数？

Question

我有一个包含很多数字的文件：

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
...

我想打印数字 0 连续重复 10 次（最少）的行数。考虑到上述输入，输出将是：4（对于行号 4，因为 0 连续重复了 10 次）。文件 list.txt 是一个巨大的文件。我是 Python 的新人。我怎样才能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)

Answer 1

逐行阅读和评估文件。如果找到模式，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

Answer 2

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

如何得到一个重复n次的数？

How obtain a number what is repeated a n times?

python

iteration

file

processing-efficiency