Python 抓取两个特定字符之间的子串

Question

我有一个包含数百个文件的文件夹，名称如下：

"2017_05_S2B_7VEG_20170528_0_L2A_B01.tif"

惯例： year_month_ID_zone_date_0_L2A_B01.tif（"_0_L2A_B01.tif"，并且 "zone" 永远不会改变）

我需要遍历每个文件并根据它们的名称构建路径以便下载它们。例如：

name = "2017_05_S2B_7VEG_20170528_0_L2A_B01.tif"
path = "2017/5/S2B_7VEG_20170528_0_L2A/B01.tif"

路径约定需要是：path = year/month/ID_zone_date_0_L2A/B01.tif

我想做一个循环，每次遇到 "_" 字符时将我的字符串“剪切”成几个部分，然后按正确的顺序拼接不同的部分以创建我的路径名。我试过了，但没用：

import re

filename = 
"2017_05_S2B_7VEG_20170528_0_L2A_B01.tif"

try:
    found = re.search('_(.+?)_', filename).group(1)
except AttributeError:
    # _ not found in the original string
    found = '' # apply your error handling

我怎样才能在 Python 上做到这一点？

Answer 1

无需正则表达式——您只需使用 split().

filename = "2017_05_S2B_7VEG_20170528_0_L2A_B01.tif"
parts = filename.split("_")

year = parts[0]
month = parts[1]

Answer 2

因为你只有一个分隔符，你不妨简单地使用 Python 的内置拆分功能：

import os

items = filename.split('_')
year, month = items[:2]
new_filename = '_'.join(items[2:])

path = os.path.join(year, month, new_filename)

Answer 3

filename = "2017_05_S2B_7VEG_20170528_0_L2A_B01.tif"
temp = filename.split('_')
result = "/".join(temp)
print(result)

结果是 2017/05/S2B/7VEG/20170528/0/L2A/B01.tif

Answer 4

试试下面的代码片段

filename = "2017_05_S2B_7VEG_20170528_0_L2A_B01.tif"
found = re.sub('(\d+)_(\d+)_(.*)_(.*)\.tif', r'///.tif', filename)
print(found) # prints 2017/05/S2B_7VEG_20170528_0_L2A/B01.tif

Answer 5

也许你可以这样做：

from os import listdir, mkdir
from os.path import isfile, join, isdir

my_path = 'your_soure_dir'

files_name = [f for f in listdir(my_path) if isfile(join(my_path, f))]

def create_dir(files_name):
    for file in files_name:
        month = file.split('_', '1')[0]
        week = file.split('_', '2')[1]
        if not isdir(my_path):
            mkdir(month)
            mkdir(week)
            ### your download code

Python 抓取两个特定字符之间的子串

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