<ul>
<li>我认为你真正的问题是如何把所有的文件放到一个数据框中</li>
<li>使用标准库的<a href="https://docs.python.org/3/library/pathlib.html" rel="nofollow noreferrer">pathlib</a>处理文件。
<ul>
<li><a href="https://realpython.com/python-pathlib/" rel="nofollow noreferrer">Python 3's pathlib Module: Taming the File System</a></li>
</ul></li>
<li>由于您的<code>csv</code>文件是相同的,如您在问题中所述,将它们全部组合到一个数据帧中,然后一次清除所有数据会更有效。
<ul>
<li>单独清理每个数据帧,然后将它们合并,效率较低</li>
</ul></li>
</ul>
<h2>得到一个单一的,组合的数据帧</h2>
<pre class="lang-py prettyprint-override"><code>from pathlib import Path
import pandas as pd
p = Path(r'c:\some_path_to_files') # set your path
files = p.glob('nba*.csv') # find your files
# It was stated, all the files are the same format, so create one dataframe
df = pd.concat([pd.read_csv(file) for file in files])
</code></pre>
<ul>
<li><code>[pd.read_csv(file) for file in files]</code>是一个列表理解,它为每个文件创建一个数据帧。你知道吗</li>
<li><code>pd.concat</code>组合列表中的所有文件</li>
</ul>
<h2>要获取单独的数据帧:</h2>
<ul>
<li>创建数据帧的<code>dict</code></li>
<li>每个<code>key</code>的<code>dict</code>将是一个文件名</li>
</ul>
<pre class="lang-py prettyprint-override"><code>p = Path(r'c:\some_path_to_files') # set your path
files = p.glob('nba*.csv') # find your files
df_dict = dict()
for file in files:
df_dict[file.stem] = pd.read_csv(file)
</code></pre>
<h3>使用<code>df_dict</code>:</h3>
<pre class="lang-py prettyprint-override"><code>df_dict.keys() # to show you all the keys
df_dict[filename] # to access a specific dataframe
# after cleaning the individual dataframes in df_dict, they can be combined
df_final = pd.concat([value for value in df_dict.values()])
</code></pre>