Python重命名文件从csv文件中读取名称

Question

嗨，我一直在尝试使 this 适应我的需求，但我只是 python 的新手，我有一个包含多列和多行的 csv 文件，重要的列是 1 = 文件的旧名称，并且2 =文件的新名称，所以我需要转到csv文件中列出的文件所在的目录并将它们重命名为第2列的新名称，正如我所说我已经尝试了很多事情但没有成功，我粘贴了最后的代码我做了，所以你有一个想法：

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)

非常感谢。

Answer 1

这将重命名每个匹配的文件，并报告任何尝试重命名的错误。它不会尝试移动不存在的文件。

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 2

您应该使用从 CSV 创建的字典 IDs：

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

但是您可能应该使用某种错误检查..（编辑正如其他答案所指出的，最好也使用os.path.join）也许是这样的：

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 3

您正在对文件进行迭代并将旧名称和新名称存储在 IDs 中，但不要使用它，而只是尝试从文件中进一步读取（这显然会失败，因为您已经通过那时）。 IOW 你应该使用你的IDs dict 来获取新名称（使用旧名称作为键），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))

现在有一个更简单的解决方案：只需在迭代 csv 文件时重命名文件即可：

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

Answer 4

这是我在使用上述 sn-ps 没有成功后想出的。您可以在没有 pandas 的情况下执行此操作，但我还需要进行格式化并从我的数据中删除一些重复值，以便 pandas 让我的生活更轻松。

import os
import pandas as pd


csv_name = "/home/your_filename.csv"
#parent_dir is the directory that has the folders you want to rename
parent_dir = "/path/to/folder"
df  =  pd.read_csv(csv_name)

list_oldnames = df['Old_Name'].astype(str).to_list()

list_newnames = df['New_Name'].astype(str).to_list()

for folder in os.listdir(parent_dir):
    for oldname, newname in zip(list_oldnames, list_newnames):
        if str(oldname) == str(folder):
            try:
                os.rename(os.path.join(parent_dir, oldname), os.path.join(parent_dir, newname))
            except:
                print("Could not rename "+ str(os.path.join(parent_dir, oldname)) )