python 遍历文本文件直到满足条件

Question

我想继续遍历文本文件，直到满足循环内的当前条件。

这里是示例文本：

10-01   N/A
10-02   N/A
10-03   N/A
10-04   N/A
10-05   N/A
10-06   N/A
10-07   N/A
10-08   N/A
10-09   N/A
10-10   N/A
10-11   N/A
10-12   N/A
===04===...... # Skip line until '01' is found
===12===...... # Skip line until '01' is found
05-01   N/A
05-02   N/A
05-03   N/A
05-04   N/A
05-05   N/A
05-06   N/A
===08===...... # Skip line until '07' is found
===11===...... # Skip line until '07' is found
05-07   N/A
05-08   N/A
05-09   N/A
05-10   N/A
05-11   N/A
05-12   N/A

这是我正在尝试的 while 循环：

x = 1
with open(loc_path + 'SAMPLEDATA.TXT', 'rb') as textin:
    for line in textin:
        while x < 13:
            if line[3:].startswith(str(x).zfill(2)):
                print '%r' % line
            else:
                x = 1 # Restart loop
            x += 1

如果使用 while 循环不正确，除了使用 while 循环之外，还有其他方法可以实现吗？

谢谢

Answer 1

您想将 while x < 13 更改为 if 语句以有条件地停止 for 循环。例如，

x = 1
with open(loc_path + 'SAMPLEDATA.TXT', 'rb') as textin:
    for line in textin:
        if line[3:].startswith(str(x).zfill(2)):
            print '%r' % line
        else:
            x = 1 # Restart counter
        x += 1
        if x >= 13:
            break # Stop reading

Answer 2

您想要：仅在找到要查找的行时增加计数器，并在达到 13 时重置它

x = 1
with open(loc_path + 'SAMPLEDATA.TXT', 'rb') as textin:
    for line in textin:
        if line[3:].startswith(str(x).zfill(2)):
            print '%r' % line
            x += 1
        if x >= 13:
            x = 1  # reset counter

python 遍历文本文件直到满足条件

python iterate though text file until condition is met

python

while-loop