如何让Python查看子文件夹？

Question

我正在尝试创建一个Python脚本，该脚本将在一系列子文件夹中查找并删除空的shapefile。 我已经成功创建了脚本部分，该部分将删除一个文件夹中的空文件，但是“项目”文件夹中总共有70个文件夹。 虽然我可以将代码复制和粘贴69次，但我确信这一定是一种方法，可以查看每个子文件夹并为每个子文件夹运行代码。 以下是我到目前为止所拥有的。 有任何想法吗？ 我对此很陌生，我只是编辑了一个现有代码就可以做到这一点。 谢谢！

import os

# Set the working directory
os.chdir ("C:/Naview/Platypus/Project")

# Get the list of only files in the current directory
file = filter(os.path.isfile, os.listdir('C:/Naview/Platypus/Project'))
# For each file in directory
for shp in file:
    # Get only the files that end in ".shp"
    if shp.endswith(".shp"):
        # Get the size of the ".shp" file.
        # NOTE: The ".dbf" file can vary is size whereas
        #       the shp & shx are always the same when "empty".
        size = os.path.getsize(shp)
        print "\nChecking " + shp + "'s file size..."

        #If the file size is greater than 100 bytes, leave it alone.                  
        if size > 100:
            print "File is " + str(size) + " bytes"
            print shp + " will NOT be deleted \n"

        #If the file size is equal to 100 bytes, delete it.      
        if size == 100:
            # Convert the int output from (size) to a string.
            print "File is " + str(size) + " bytes"                    
            # Get the filename without the extention
            base = shp[:-4]
            # Remove entire shapefile
            print "Removing " + base + ".* \n"
            if os.path.exists(base + ".shp"):
               os.remove(base + ".shp")
            if os.path.exists(base + ".shx"):
                os.remove(base + ".shx")
            if os.path.exists(base + ".dbf"):
                os.remove(base + ".dbf")
            if os.path.exists(base + ".prj"):
                os.remove(base + ".prj")
            if os.path.exists(base + ".sbn"):
                os.remove(base + ".sbn")
            if os.path.exists(base + ".sbx"):
                os.remove(base + ".sbx")
            if os.path.exists(base + ".shp.xml"):
                os.remove(base + ".shp.xml")

Answer 1

有几种方法可以做到这一点。 我是glob的粉丝

for shp in glob.glob('C:/Naview/Platypus/Project/**/*.shp'):
    size = os.path.getsize(shp)
    print "\nChecking " + shp + "'s file size..."

    #If the file size is greater than 100 bytes, leave it alone.                  
    if size > 100:
        print "File is " + str(size) + " bytes"
        print shp + " will NOT be deleted \n"
        continue
    print "Removing", shp, "files"
    for file in glob.glob(shp[:-3] + '*'):
        print " removing", file
        os.remove(file)

Answer 2

是时候学习过程编程了： 定义函数 。

将您的代码放入带有path参数的函数中，并为70条路径中的每条路径调用它：

def delete_empty_shapefiles(path):
    # Get the list of only files in the current directory
    file = filter(os.path.isfile, os.listdir(path))
    ...

paths = ['C:/Naview/Platypus/Project', ...]
for path in paths:
    delete_empty_shapefiles(path)

奖励点，用于创建执行os.path.exists()和os.remove()调用的函数。