Python抓取两个特定字符之间的子字符串

Question

我有一个文件夹，里面有几百个文件，命名为:

"2017_05_S2B_7VEG_20170528_0_L2A_B01.tif"

公约:year_month_ID_zone_date_0_L2A_B01.tif("_0_L2A_B01.tif","zone"不变)

我需要的是遍历每个文件，并根据它们的名称构建一个路径，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"

路径约定为:path = year/month/ID_zone_date_0_L2A/B01.tif

我想做一个循环，可以"切割";每次遇到"_"字符时，将字符串分成几个部分，然后按正确的顺序将不同部分拼接起来，以创建我的路径名。我试过这个，但它没有工作:

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

如何在Python上实现呢?

Answer 1

因为你只有一个分隔符，你也可以直接使用Python内置的split函数:

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

Answer 2

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'1/2/3/4.tif', filename)
print(found) # prints 2017/05/S2B_7VEG_20170528_0_L2A/B01.tif

Answer 3

不需要正则表达式——您可以直接使用split()。

filename = "2017_05_S2B_7VEG_20170528_0_L2A_B01.tif"
parts = filename.split("_")
year = parts[0]
month = parts[1]

Answer 4

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

Answer 5

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

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