擅长:python、mysql、java
<p>我认为您可以使用<a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.setdiff1d.html" rel="nofollow noreferrer">^{<cd1>}</a>和<a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.in1d.html" rel="nofollow noreferrer">^{<cd2>}</a>并按<a href="http://pandas.pydata.org/pandas-docs/stable/indexing.html#boolean-indexing" rel="nofollow noreferrer">^{<cd3>}</a>过滤:</p>
<pre><code>diffs = np.setdiff1d(price_5min.index.date, price.index.date))
df = price_5min[~np.in1d(price_5min.index.date, diffs]
</code></pre>
<p>另一个带有<a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DatetimeIndex.floor.html" rel="nofollow noreferrer">^{<cd4>}</a>或<a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DatetimeIndex.to_period.html" rel="nofollow noreferrer">^{<cd5>}</a>的解决方案:</p>
^{pr2}$
<hr/>
<pre><code>dates = price.index.to_period('D')
dates_5min = price_5min.index.to_period('D')
df = price_5min[~dates_5min.isin(dates_5min.difference(dates))]
</code></pre>