【问题标题】:Python: Replace one list with another list?Python:用另一个列表替换一个列表?
【发布时间】:2013-08-26 15:57:37
【问题描述】:

我被困住了。我正在将我们网络上的文件夹移动到一个中心位置,这些文件夹都有一个唯一的 ID。有一些文件夹有拼写错误,因此与中心位置的唯一 ID 不匹配。我找到了正确的 IDS,但我需要在移动它们之前重命名这些文件夹。例如,我创建了一个具有错误唯一 ID 的 Excel 电子表格,并且在单独的列中具有正确的 ID。现在,我想用正确的 ID 重命名文件夹,然后将这些文件夹转移到中心位置。我的代码......粗糙,因为我想不出一个好的方法来做到这一点。我觉得使用列表是要走的路,但是由于我的代码正在遍历文件夹,所以我不确定如何实现这一点

编辑:我认为像 this 这样的东西可能是我正在寻找的东西

例如: 在文件夹 A 中:一个名为 12334 的文件应该重命名为 1234。然后移动到文件夹 1234 中的基本目录。

这是我的代码:

import os
import re
import sys
import traceback
import collections
import shutil

movdir = r"C:\Scans"
basedir = r"C:\Links"
subfolder = "\Private Drain Connections"

try:
    #Walk through all files in the directory that contains the files to copy
    for root, dirs, files in os.walk(movdir):
        for filename in files:
            #find the name location and name of files
            path = os.path.join(root, filename)

            #file name and extension
            ARN, extension = os.path.splitext(filename)
            print ARN

            #Location of the corresponding folder in the new directory
            link = os.path.join(basedir, ARN)
            if not os.path.exists(link):
                newname = re.sub(372911000002001,372911000003100,ARN)
                newname =re.sub(372809000001400,372909000001400,ARN)
                newname =re.sub(372809000001500,372909000001500,ARN)
                newname =re.sub(372809000001700,372909000001700,ARN)
                newname = re.sub(372812000006800,372912000006800,ARN)
                newname =re.sub(372812000006900,372912000006900,ARN)
                newname =re.sub(372812000007000,372912000007000,ARN)
                newname =re.sub(372812000007100,372912000007100,ARN)
                newname =re.sub(372812000007200,372912000007200,ARN)
                newname =re.sub(372812000007300,372912000007300,ARN)
                newname =re.sub(372812000007400,372912000007400,ARN)
                newname =re.sub(372812000007500,372912000007500,ARN)
                newname =re.sub(372812000007600,372912000007600,ARN)
                newname =re.sub(372812000007700,372912000007700,ARN)
                newname =re.sub(372812000011100,372912000011100,ARN)


                os.rename(os.path.join(movdir, ARN, extension ),
                os.path.join(movdir, newname, extension))
                oldpath = os.path.join(root, newname)
                print ARN, "to", newname
                newpath = basedir + "\\" + newname + subfolder
                shutil.copy(oldpath, newpath)
                print "Copied"

except:
    print ("Error occurred")

感谢下面的答案,这是我的最终代码:

import arcpy
import os
import re
import sys
import traceback
import collections
import shutil

movdir = r"C:\Scans"
basedir = r"C:\Links"
subfolder = "\Private Drain Connections"


import string

l = ['372911000002001',
'372809000001400',
'372809000001500',
'372809000001700',
'37292200000800'
]
l2 = ['372911000003100',
'372909000001400',
'372909000001500',
'372909000001700',
'372922000000800'
]

try:
    #Walk through all files in the directory that contains the files to copy
    for root, dirs, files in os.walk(movdir):
        for filename in files:
            #find the name location and name of files
            path = os.path.join(root, filename)

            #file name and extension
            ARN, extension = os.path.splitext(filename)
            oldname = str(ARN)
            #Location of the corresponding folder in the new directory
            link = os.path.join(basedir, ARN)

            if not os.path.exists(link):
                for ii, jj in zip(l, l2):
                    newname = re.sub(ii,jj, ARN)
                    newname = str(newname)
                    print path
                    newpath = os.path.join(root, oldname) +  extension
                    print "new name", newpath
                    os.rename(path, newpath)
                    print "Renaming"
                    newpath2 = basedir + "\\" + newname + subfolder
                    shutil.copy(newpath, newpath2)
                    print "Copied"
                if newname != ARN:
                    break


            else:
                continue

except:
    print ("Error occurred")
    tb = sys.exc_info()[2]
    tbinfo = traceback.format_tb(tb)[0]
    pymsg = "PYTHON ERRORS:\nTraceback Info:\n" + tbinfo + "\nError Info:\n    " + \
        str(sys.exc_type)+ ": " + str(sys.exc_value) + "\n"
    msgs = "GP ERRORS:\n" + arcpy.GetMessages(2 )+ "\n"
    print (pymsg)
    print (msgs)

【问题讨论】:

  • 这...不仅仅是替换一个列表...
  • @IgnacioVazquez-Abrams 我的代码的另一部分已关闭并可以正常工作,我只需要一种更有效的方法来替换坏名。
  • 高效还是优雅?我们在谈论多少错误的文件夹?
  • @Jblasco 目前只有 30 个
  • 那我不认为效率会成为问题……我认为优雅优先;)

标签: python list rename


【解决方案1】:

一个想法:尝试将 id 转换为字符串。我的意思是:

            newname = re.sub('372911000002001','372911000003100',ARN)

希望对你有帮助!

【讨论】:

    【解决方案2】:

    对我来说,要走的路是将两个列表都读入列表对象:

    list1 = ["372911000002001", "372809000001400", "372809000001500"]
    list2 = ["372911000003100", "372909000001400", "372909000001500"]
    for ii, jj in zip(list1, list2):
        newname = re.sub(ii,jj,ARN)  #re.sub returns ARN if no substitution done
        if newname != ARN:
           break
    

    【讨论】:

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