有一个由人和订单组成的数据框......
person order elements
Alice [drink, snack, salad, fish, dessert] 5
Tom [drink, snack] 2
John [drink, snack, soup, chicken] 4
Mila [drink, snack, soup] 3
我想知道顾客的主餐是什么。因此，我想添加另一列 [main_meal] 以便它成为我的 df。
person order elements main_meal
Alice [drink, snack, salad, fish, dessert] 5 fish
Tom [drink, snack] 2 none
John [drink, snack, soup, chicken] 4 chicken
Mila [drink, snack, soup] 3 none
规则是，如果客户点了 4 餐或更多餐，则意味着第 4 个元素始终是主菜，所以我想从订单列的列表中提取第 4 个元素。如果它包含的元素少于 4 个，则将 'main_meal' 分配给 none。我的代码:
df['main_meal'] = ''
if df['elements'] >= 4:
df['main_meal'] = df.order[3]
else:
df['main_meal'] = 'none'
它不起作用:
ValueError Traceback (most recent call last)
<ipython-input-100-39b7809cc669> in <module>()
1 df['main_meal'] = ''
2 df.head(5)
----> 3 if df['elements'] >= 4:
4 df['main_meal'] = df.order[3]
5 else:
~\Anaconda\lib\site-packages\pandas\core\generic.py in __nonzero__(self)
1571 raise ValueError("The truth value of a {0} is ambiguous. "
1572 "Use a.empty, a.bool(), a.item(), a.any() or
a.all()."
-> 1573 .format(self.__class__.__name__))
1574
1575 __bool__ = __nonzero__
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
我的代码有什么问题？
请您参考如下方法:
使用str方法切片
In [324]: df['order'].str[3]
Out[324]:
0 fish
1 NaN
2 chicken
3 NaN
Name: order, dtype: object
In [328]: df['main_meal'] = df['order'].str[3].fillna('none')
In [329]: df
Out[329]:
person order elements main_meal
0 Alice [drink, snack, salad, fish, dessert] 5 fish
1 Tom [drink, snack] 2 none
2 John [drink, snack, soup, chicken] 4 chicken
3 Mila [drink, snack, soup] 3 none
