Question

我有大量文件存储在文件夹中。我需要对每个文件对执行相同的python脚本，并将结果输出到txt或excel文件中。如何自动化进程python或批处理脚本？
例如。文件夹中的文件

ap.hdf5
sta.hdf5
ap_20150909_154518_00.hdf5
sta_20150909_154518_00.hdf5
ap_20150909_154530_00.hdf5
sta_20150909_154530_00.hdf5
ap_20150909_154541_00.hdf5
sta_20150909_154541_00.hdf5

这些文件是根据修改的数据排列的。我需要对每个对执行相同的python脚本并在文本文件中输出结果例如

python result.py ap.hdf5 sta.hdf5
python result.py ap_20150909_154518_00.hdf5 sta_20150909_154518_00.hdf5

如何创建可以自动执行流程的批处理文件？提前谢谢。

编辑：文件夹内的文件略有不同。

ap.hdf5
sta.hdf5
ap_20150909_154518_00.hdf5
sta_20150909_154524_00.hdf5
ap_20150909_154530_00.hdf5
sta_20150909_154536_00.hdf5
ap_20150909_154541_00.hdf5
sta_20150909_154547_00.hdf5

这里，在AP.hdf5

之后的几秒钟记录Sta文件

Answer 1

您可以使用glob module获取以 ap 开头的文件列表。然后，您可以将 ap 更改为 sta 以获取配对文件的名称（假设始终存在一对文件）。如果你有这个，你可以处理它们（就像你之前做的那样）。

import glob
# iterate over all files starting with ap and ending in .hdf5
for file_a in glob.iglob("ap*.hdf5"):
    # replace the beginning of the filename with sta
    file_b = "sta" + file_a[2:]

    # do your processing (result.py) using file_a and file_b as your pair

Answer 2

如果您真的不想修改result.py来执行此操作，可以使用以下方法：

import subprocess
import glob
import os.path

with open('output.txt', 'w') as f_output:
    # ap files sorted by modified order
    files = sorted(glob.glob(r'ap*.hdf5'), key=lambda x: os.path.getmtime(x))

    for ap in files:
        path, filename = os.path.split(ap)
        sta = os.path.join(path, 'sta{}'.format(filename[2:]))

        # Do we have an ap/sta pair?
        if os.path.exists(sta):
            # Launch the Python script with the required parameters
            p = subprocess.Popen(['python.exe', 'result.py', ap, sta], stdout=subprocess.PIPE, stderr=subprocess.PIPE)
            out, err = p.communicate()
            # Write stdout to a file and stderr to the screen
            f_output.write(out)
            print err
        else:
            print '{} is missing'.format(sta)

这将按照修改日期的顺序为每个ap sta文件对运行Python脚本，并将脚本中的任何输出写入output.txt

更新 - 根据更新的问题，以下脚本将根据文件名配对ap和sta个文件。如果找不到合适的一对，它将停止：

import subprocess
import glob
import os.path
import itertools

def sort_by_ending(filename):
    filename = os.path.split(filename)[1]
    if '_' in filename:
        return filename.split('_')[1:]
    else:
        return [filename]

folder = r'c:\test'

with open('output.txt', 'w') as f_output:
    # ap and sta files sorted by filename ending
    files = sorted(glob.glob(os.path.join(folder, 'ap*.hdf5')) + glob.glob(os.path.join(folder, 'sta*.hdf5')), key=sort_by_ending)

    for ap, sta in itertools.izip(*[iter(files)]*2):
        print "'{}' and '{}'".format(os.path.split(ap)[1], os.path.split(sta)[1])
        # Do we have an ap/sta pair?
        if os.path.split(ap)[1].startswith('ap') and os.path.split(sta)[1].startswith('sta'):
            # Launch the Python script with the required parameters
            p = subprocess.Popen(['python.exe', 'result.py', ap, sta], stdout=subprocess.PIPE, stderr=subprocess.PIPE)
            out, err = p.communicate()
            # Write stdout to a file and stderr to the screen
            f_output.write(out)
            print err
        else:
            print '{} is missing'.format(sta)
            break

对于您给定的文件名示例，这将打印以下输出：

'ap_20150909_154518_00.hdf5' and 'sta_20150909_154524_00.hdf5'

'ap_20150909_154530_00.hdf5' and 'sta_20150909_154536_00.hdf5'

'ap_20150909_154541_00.hdf5' and 'sta_20150909_154547_00.hdf5'

'ap.hdf5' and 'sta.hdf5'

Answer 3

import glob, os

ap_files = glob.glob("outdir/ap*.hdf5") #use glob to get ap files
for ap_file in ap_files:   #walk through the ap file names
  basename = os.path.basename(ap_file)  #split of the base name
  sta_file = "outdir/" + basename.replace("ap","sta",1) #make the sta name
  dothingsto(ap_file, sta_file)  #do whatever you wish to the two files

对按日期排列的文件夹中的文件重复执行python脚本

3 个答案: