Python 读取.TXT文件，如果值为真，则打印响应

我有一个像这样的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是否可以读取此值并检测值中的大间隙，请参见检测11和12数字从27.30大幅跳到147.91，脚本是否可以读取此值并打印响应，例如“检测到间隙”以及持续时间

---

谢谢

首先，我需要模拟您的数据以模拟文件处理程序

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，我需要提取起始值和结束值：

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]

的重要假设是字符串的格式始终正确。你可以看到我将其转换为浮动，这样我就可以将它们作为数字来处理

接下来，我将这些值与任意阈值进行比较（这里是

20

）：

输出：

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

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

---

使用整个数组的平均值定义大步进

我这样解决了问题：

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

代码向我们展示了这个输出：

re.findall

使用正则表达式从字符串中提取浮点数

---

希望这有帮助

如果你的问题只是“有可能吗？”，那么答案是肯定的。它甚至没有那么难。你能告诉我们到目前为止你试过什么吗？作为一个快速提示，我建议您看看python中的

readline（）

，它允许您逐行读取文件。当您在线时，您可以访问所需的数据

147.91

，并对照上一个数据

27.30

（如果存在）进行检查。希望这有助于Irek，问题是，虽然我尝试了readline，但每个txt文件中的值都不同，所以我丢失了，因为它不是我每次都要查找的特定值。我会尝试在行中fin

start:

和

end:

，并提取这些值，这是完全可行的。因此，只要这些子字符串在数值之前，就可以提取它们。除非你说这条线可能是这样的：

[2019-04-03 13:59:15289]|[DET]：检测12（开始：147.91，完成：149.09）

？

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