读取.TXT文件并打印响应（如果值为true）

Question

我有一个看起来像这样的TXT文件

[2019-04-03 13:59:15,034] | [DET ]: Detection 1 (start: 0.83, end: 1.57)
[2019-04-03 13:59:15,044] | [DET ]: Detection 2 (start: 1.74, end: 6.74)
[2019-04-03 13:59:15,062] | [DET ]: Detection 3 (start: 6.74, end:11.74)
[2019-04-03 13:59:15,071] | [DET ]: Detection 4 (start:11.74, end:15.97)
[2019-04-03 13:59:15,072] | [DET ]: Detection 5 (start:16.06, end:18.61)
[2019-04-03 13:59:15,081] | [DET ]: Detection 6 (start:18.82, end:20.60)
[2019-04-03 13:59:15,090] | [DET ]: Detection 7 (start:20.61, end:22.46)
[2019-04-03 13:59:15,094] | [DET ]: Detection 8 (start:22.48, end:23.58)
[2019-04-03 13:59:15,099] | [DET ]: Detection 9 (start:23.75, end:25.61)
[2019-04-03 13:59:15,101] | [DET ]: Detection 10 (start:25.64, end:26.60)
[2019-04-03 13:59:15,102] | [DET ]: Detection 11 (start:26.61, end:27.30)
[2019-04-03 13:59:15,289] | [DET ]: Detection 12 (start:147.91, end:149.09)
[2019-04-03 13:59:15,289] | [DET ]: Detection 13 (start:149.10, end:154.10)
[2019-04-03 13:59:15,289] | [DET ]: Detection 14 (start:154.10, end:156.03)

Python是否有可能读取此值并检测到较大的值差距，请参见检测11和12，从27.30到147.91的数字有较大的跳跃，脚本可以读取此值并打印响应，例如“持续时间？

谢谢！

Answer 1

首先，我需要模拟您的数据以模仿文件处理程序。

s = """[2019-04-03 13:59:15,034] | [DET ]: Detection 1 (start: 0.83, end: 1.57)
# Omitted for brevity
[2019-04-03 13:59:15,102] | [DET ]: Detection 11 (start:26.61, end:27.30)
[2019-04-03 13:59:15,289] | [DET ]: Detection 12 (start:147.91, end:149.09)
[2019-04-03 13:59:15,289] | [DET ]: Detection 13 (start:149.10, end:154.10)
[2019-04-03 13:59:15,289] | [DET ]: Detection 14 (start:154.10, end:156.03)""".split("\n")

接下来，我需要提取开始和结束值：

starts = []
ends = []
for line in s:
  # Extract start val
  start = line.split("start:")[1].strip().split(",")[0]
  # Cast as float
  start = float(start)
  starts.append(start)

  # Extract end val
  end = line.split("end:")[1].strip().split(")")[0]
  # Cast as float
  end = float(end)
  ends.append(end)

请注意，end = line.split("end:")[1].strip().split(")")[0]的重要假设是您的字符串应始终正确格式化。而且您看到我被迫漂浮，所以我可以将它们当作数字使用。

接下来，我将这些值与任意阈值进行比较（此处为20）：

for i in range(1,len(starts)):
  if starts[i]-ends[i-1] > 20:
    print("Gap detected: {}".format(s[i]))
  else:
    pass

输出：

Gap detected: [2019-04-03 13:59:15,289] | [DET ]: Detection 12 (start:147.91, end:149.09)

Answer 2

代码：

import re
import statistics as s
with open('name_of_the_file.txt','r') as f:
    z = f.readlines()
data = []
for i in z:
    temp = i.split('end')
    m =  re.findall(r"[-+]?\d*\.\d+|\d+", temp[1])
    if  m != None:
        data.append(float(m[0]))
    else:
        print(i)
for j in range(len(data)-1):
    cal = abs(data[j] - data[j+1])
    if ( cal > s.mean(data)):
        print(cal)
        print('Gap detected at '+str(j+2))

输出：

121.79
Gap detected at 12

使用整个数组的均值定义较大步长

Answer 3

我这样解决了这个问题：

import re

previous_end = -1

# Reading a file
with open('file.txt') as file:
    # Iterating over each line
    for line in file:
        # Spliting on ":"
        arr_lin = line.split(":")
        # Very bad way of accessing the number just after 'start' and 'end' 
        # However, if the file structure is always identical, so "(start: 0.83, end: 1.57)" this will do the job
        # Make sure we access the right index and convert it to float
        start = float(re.findall(r"[-+]?\d*\.\d+|\d+", arr_lin[-2])[0])
        end = float(re.findall(r"[-+]?\d*\.\d+|\d+", arr_lin[-1])[0])
        # Checking for your gap
        if (previous_end != -1):
            print("Gap detected {}".format(start-previous_end))
        previous_end = end

代码向我们显示了以下输出：

re.findall使用正则表达式从字符串中提取浮点数。

希望有帮助