您已接近，需要使用str进行索引，该索引适用于Serie s的每个值：

data['Order_Date'] = data['Shipment ID'].str[:8]

如果没有NaN s值，则为了获得更好的性能：
x一个一个一个一个x一个一个二个x
如果忽略str代码，则按位置过滤列，前N个值如下：

print (data['Shipment ID'][:2])
0    20180504-S-20000
1    20180514-S-20537
Name: Shipment ID, dtype: object

您也可以使用str.extract

• • 例如：**

import pandas as pd

df = pd.DataFrame({'Shipment ID': ['20180504-S-20000', '20180514-S-20537', '20180514-S-20541', '20180514-S-20644', '20180514-S-20644', '20180516-S-20009', '20180516-S-20009', '20180516-S-20009', '20180516-S-20009']})
df["Order_Date"] = df["Shipment ID"].str.extract(r"(\d{8})")
print(df)

• • 输出：**

Shipment ID Order_Date
0  20180504-S-20000   20180504
1  20180514-S-20537   20180514
2  20180514-S-20541   20180514
3  20180514-S-20644   20180514
4  20180514-S-20644   20180514
5  20180516-S-20009   20180516
6  20180516-S-20009   20180516
7  20180516-S-20009   20180516
8  20180516-S-20009   20180516

您还可以决定从-S删除到末尾

df["Order_Date"]=df['Shipment ID'].replace(regex=r"\-.*",value="")
df
        Shipment ID Order_Date
0  20180504-S-20000   20180504
1  20180514-S-20537   20180514
2  20180514-S-20541   20180514
3  20180514-S-20644   20180514
4  20180514-S-20644   20180514
5  20180516-S-20009   20180516
6  20180516-S-20009   20180516
7  20180516-S-20009   20180516
8  20180516-S-20009   20180516

您还可以捕获前8位数字，然后删除所有内容，并使用捕获组的反向引用替换：

df['Shipment ID'].replace(regex=r"(\d{8}).*",value="\\1")

我决定使用带有子集的.str访问器：

data['p_dt'] = data['timestamp'].astype(str).str[:8]