python - 读取文件名，对其进行操作然后保存

Question

基本上我正在读取一个名为 Sounding 的文件，它的名称为“12142014_2345.csv”，我想将其保存为

'12142014_2345_Averaged.csv'

下面是我引导它的代码。

basename = os.path.basename(Sounding)
basename,ext = os.path.split(basename)
    with open(os.path.join(basename+'_Averaged'+ext)) as f:
           w = csv.DictWriter(f, rows_1[0].keys())
           w.writeheader()

这是我的错误。

Traceback (most recent call last):
  File "C:\Users\Bud\Desktop\OWLES RECENT\Moving Average.py", line 151, in <module>
    with open(os.path.join(basename+'_Averaged'+ext)) as f:
IOError: [Errno 2] No such file or directory: u'_AveragedUIllinois_20131207Singular.csv'

我不确定我做错了什么。

Answer 1

您想使用 os.path.splitext (拆分扩展名)而不是 split (分割最后一个路径元素)。不要忘记以写入模式打开文件(并检查缩进):

basename,ext = os.path.splitext(basename)
with open(os.path.join(basename+'_Averaged'+ext), 'w') as f:
       w = csv.DictWriter(f, rows_1[0].keys())
       w.writeheader()