Python重命名文件從csv文件讀取名稱

Question

您好，我一直在努力適應這個我需要，但我只是在python newbe，我有多個列和行的csv文件，重要的列1 =文件的舊名稱，以及2 =新名稱文件，因此我需要轉到csv文件中列出的文件所在的目錄，並將其重命名為第2列的新名稱，因為我說我嘗試了很多事情都沒有成功，所以我粘貼了我所做的最后一個代碼你有一個主意：

import os, unicodecsv as csv, sys

IDs = {}

#open and store the csv file
with open('documentos_corpus_ladino.csv','rb') as csvfile:
        timeReader = csv.reader(csvfile, delimiter = ',')

        # build a dictionary with the associated IDs
        for row in timeReader:
              IDs[ row[0] ] = row[1]

# #get the list of files
path = 'txt_orig/'
tmpPath = 'txt_tmp/'
for filename in os.listdir('txt_orig/'):
    oldname = filename
    newname = filename.replace(oldname, csvfile.next().rstrip().split(",")[1])
    os.rename(path + filename, tmpPath + newname)

非常感謝。

Answer 1

您應該使用從CSV創建的字典IDs ：

for filename in os.listdir(path):
    oldname = filename
    newname = IDs[oldname]
    os.rename(path + filename, tmpPath + newname)

但是您可能應該使用某種錯誤檢查。.（ 編輯如其他答案所指出，最好也使用os.path.join ）也許遵循以下原則：

failed = []
for oldname in os.listdir(path):
    try:
        old = os.path.join(path, oldname)
        new = os.path.join(tmpPath, IDs[oldname])
        os.rename(old, new)
    except KeyError, OSError:
        failed.append(oldname)

print failed

Answer 2

這將重命名每個匹配的文件，並報告任何嘗試重命名的錯誤。 它不會嘗試移動不存在的文件。

import os, unicodecsv as csv
# open and store the csv file
IDs = {}
with open('documentos_corpus_ladino.csv','rb') as csvfile:
    timeReader = csv.reader(csvfile, delimiter = ',')
    # build dictionary with associated IDs
    for row in timeReader:
        IDs[row[0]] = row[1]
# move files
path = 'txt_orig/'
tmpPath = 'txt_tmp/'
for oldname in os.listdir(path):
    # ignore files in path which aren't in the csv file
    if oldname in IDs:
        try:
            os.rename(os.path.join(path, oldname), os.path.join(tmpPath, IDs[oldname]))
        except:
            print 'File ' + oldname + ' could not be renamed to ' + IDs[oldname] + '!'

Answer 3

您正在文件上進行迭代，並將舊名稱和新名稱存儲在IDs但不使用它，而只是嘗試從文件中進一步讀取（這顯然會失敗，因為此時您已經讀取了整個文件）。 IOW，您應該使用IDs字典來獲取新名稱（使用舊名稱作為鍵），即：

path = 'txt_orig' # no trailing slash required
tmpPath = 'txt_tmp' # idem
for filename in os.listdir(path):
    try:
       newname = IDs[filename]
    except KeyError:
       print "no new name for '%s'" % filename
       continue
    else:     
        os.rename(os.path.join(path, filename), os.path.join(tmpPath, newname))

現在有一個更簡單的解決方案：在對csv文件進行迭代時，只需重命名文件即可：

path = 'txt_orig'
tmp_path = 'txt_tmp'

with open('documentos_corpus_ladino.csv','rb') as csvfile:
    reader = csv.reader(csvfile, delimiter = ',')
    for row in reader:
       oldname = os.path.join(path, row[0])
       if os.path.exists(oldname):
           newname = os.path.join(tmp_path, row[1])
           os.rename(oldname, newname)
           print >> sys.stderr, "renamed '%s' to '%s'" % (oldname, newname)
       else:
           print >> sys.stderr, "file '%s' not found" % oldname