擅长:python、mysql、java
<p>假设您的示例中有一个输入错误(99是98)。您可以执行以下操作:</p>
<pre><code>df = pd.DataFrame([["2020-01-01",24,123],
["2020-01-10",24,145],
["2020-01-01",58,89],
["2020-01-10",58,67],
["2020-01-01",98,34],
["2020-01-10",98,86],
["2020-01-10",67,140],
["2020-01-10",32,321],
["2020-01-10",75,76]],columns = ["date","customer_id","amount_spent" ])
df["order"] = df.groupby("customer_id").cumcount()
df[(df["date"] == "2020-01-10") & (df["order_x"]==0)]
</code></pre>
<p>输出:</p>
<pre><code> date customer_id amount_spent order_x order_y
6 2020-01-10 67 140 0 0
7 2020-01-10 32 321 0 0
8 2020-01-10 75 76 0 0
</code></pre>
<p>这需要根据df的复杂性进行编辑</p>