将子文件夹名与文件名和输出文件夹python匹配
将子文件夹名与文件名和输出文件夹python匹配
提问于 2020-10-22 15:36:50
我有以下文件夹文件结构：
- mainfolder_segment_polygon
- folder_poly5numSeg
- subfolder_compactness40
- subfolder_aoi1
- file_aoi1_seg0.shp
- file_aoi1_seg1.shp
- subfolder_aoi1
- subfolder_compactness40
- folder_poly5numSeg
- subfolder\_aoi2
- file\_aoi2\_seg0.shp
- file\_aoi2\_seg1.shp- folder\_poly6numSeg
- subfolder\_compactness40
- subfolder\_aoi1
- file\_aoi1\_seg0.shp
- file\_aoi1\_seg1.shp - subfolder\_aoi2
- file\_aoi2\_seg0.shp
- file\_aoi2\_seg1.shp我希望能够从同一个文件夹(segment_polygon)加载所有文件，向它们应用一个函数，并将其导出到具有相同结构的另一组文件夹(segment_multipoly)。
r".\segmentation_aoi\segment_multipoly\multi5numSeg\compactness40\aoi1"
- 来自
r".\segmentation_aoi\segment_polygon\poly5numSeg\compactness40\aoi1"的文件应该一起处理并导出到r".\segmentation_aoi\segment_polygon\poly5numSeg\compactness40\aoi1"
r".\segmentation_aoi\segment_multipoly\multi6numSeg\compactness40\aoi2"
- 来自
r".\segmentation_aoi\segment_polygon\poly6numSeg\compactness40\aoi2"的文件应该一起处理并导出到r".\segmentation_aoi\segment_polygon\poly6numSeg\compactness40\aoi2"
等等..。
名称“主文件夹”、“文件夹”、“子文件夹”、“文件”只是为了指示名称属于哪个级别，但它们不是文件夹标签的一部分。
input_path = os.path.join(src, "segment_polygon\\")
output_path = os.path.join(src, "segment_multipoly\\")
root = Path(input_path)
for maindir, subdirs, shpfiles in os.walk(input_path):
for shp in shpfiles:
aoi_root, shp_ext = shp.split("_")
for file in root.glob("*/*/*/*.shp"):
part_path = Path(file).parts
folder_numSeg_name = part_path[9] #here I get the subfolder "poly5numSeg", "poly6numSeg", etc
folder_aoi_name = part_path[11] #here I get the subfolder "aoi1", "aoi2", etc...
aoiprep_seg = part_path[12] # here I get the name of the file "aoi1_seg0.shp", aoi1_seg1.shp", etc
if aoi_root == folder_aoi_name:
'''apply a function to shp'''
shp.to_file(os.path.join(output_path, folder_numSeg_name, "compactness40\\", folder_aoi_name, shp)我有点迷路了。工作在Windows 10，Python 3。谢谢您的帮助。
更新脚本
segment_polygon = os.path.join(output, "segment_polygon\\") # input path
segment_multipoly = os.path.join(output, "segment_multipoly\\") # output path
# 1. get aoi directories
aoi_dir = [path for path in glob.glob(os.path.join(segment_polygon, "*/*/*"))
if os.path.isdir(path)]
# list to store the shapefiles to be intersected
input_list = []
for path in aoi_dir:
# 2. get the files
shp_paths = glob.glob(path + os.sep + '*.shp')
for shp_path in shp_paths:
# 3. do things with shp_path
full_path, seg_shp = os.path.split(shp_path)
aoi_folder = full_path[-5:] # aoi01, aoi02, aoi03....aoi25
if seg_shp.startswith(aoi_folder):
input_list.append(shp_path) # creates the new list with shapefiles that start with the same aoiX value
auto_inter = gpd.GeoDataFrame.from_file(input_list[0]) #process shp
for i in range(len(input_list)-1):
mp = gpd.GeoDataFrame.from_file(input_list[i+1]) # process shp
auto_inter = gpd.overlay(auto_inter, mp, how='intersection') #process shp
print(f"shp included in the list:\n {input_list}")
# 4. create your output file path
print(full_path)
output_path = full_path.replace("poly", "multi")
N_output_path = output_path.replace("gon", "polygon")
print(f"output_path:\n {N_output_path}")
# make sure the directories exist
if not os.path.exists(os.path.dirname(N_output_path)):
os.makedirs(os.path.dirname(N_output_path), exist_ok=True)
# create output file name
multipoly_name = aoi_folder + ".shp"
# export
auto_inter.to_file(os.path.join(N_output_path, multipoly_name)) #export shp合并了来自ygorg的变更。然而，它需要所有的形状文件的交叉口。我只想要交交和保存在aoi1文件夹上的aoi1文件。然后，aoi2格式文件并保存到aoi2文件夹中，等等。这还不起作用。
回答 2
Stack Overflow用户
发布于 2020-10-22 15:52:01
将os.walk和glob混在一起似乎相当令人困惑。如果您想处理每个aoiX文件夹。尝试首先列出所有这些目录，然后列出每个目录中的.shp文件，然后应用该函数，最后创建output_path并将其写入。
在处理文件时，分解所需的内容以避免不堪重负总是很好的。
# 1. get aoi directories
aoi_dir = [path for path in glob.glob('segment_polygon/*/*/*')
if os.path.isdir(path)]
for path in aoi_dir:
# 2. get the files
shp_paths = glob.glob(path + os.sep + '*.shp')
for shp_path in shp_paths:
# 3. do things with shp_path
# 4. create your output file path
output_path = shp_path.replace('segment_polygon', 'segment_multipoly')
# make sure the directories exist
os.makedirs(os.path.dirname(output_path), exist_ok=True)
# write in output file并且总是在没有处理或写任何东西的情况下进行一次模拟运行，并且打印路径，这样您就可以确定会发生什么了！
Stack Overflow用户
发布于 2020-10-29 16:26:56
我设法解决了这个问题。谢谢ygorg的投入。它把我引向了正确的道路。
# Create a list of the subfolders of segment_polygon
poly_dir = [path for path in glob.glob(os.path.join(segment_polygon, "*/*/*"))
if os.path.isdir(path)]
for aoi_poly in poly_dir:
# define input folder
input_subfolder = aoi_poly.split("segment_polygon\\")[1] # splits the path at "...\\" and keeps the tail (position:1)
#print(f"input folder: {input_subfolder}")
#define export folder
export_subfolder = input_subfolder.replace("poly", "multi")
export_folder = os.path.join(segment_multipoly, export_subfolder)
#print(f"output folder: {export_folder}")
# define name output shapefile
numseg, compactness, aoi = [int(s) for s in re.findall(r'\d+', aoi_poly)] #extract only the integers from the "poly" path
name_output = "aoi" + str(aoi)+ "_" + "numSeg"+ str(numseg) + "_c" + str(compactness) + ".shp" # str() is used to concatenate integers as part of the string
#print(f"shapefile label: {name_output}")
full_outputpath = os.path.join(export_folder, name_output)
#print(f"full output path: {full_outputpath}")
# intersect and merge all single polygons
input_list = list(filter(lambda mpoly: mpoly.endswith('.shp'), os.listdir(aoi_poly)))
###### apply my function here ######
# export
filetoexport.to_file(full_outputpath)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/64485683
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