根据字母"l"或"L"是否在另一列的字符串中创建新列

问题描述

我正在使用Open Food Facts数据集，该数据集非常混乱.有一个称为数量的列，其中包含有关相应食物数量的信息.条目看起来像:

I am working with the Open Food Facts dataset which is very messy.There is a column called quantity in which in information about the quantity of respective food.the entries look like:

365 g (314 ml)
992 g
2.46 kg
0,33 litre
15.87oz
250 ml
1 L
33 cl

...等等(非常混乱！)我想创建一个名为is_liquid的新列.我的想法是，如果数量字符串包含l或L，则此行中的is_liquid字段应为1，否则为0.这是我尝试过的:我写了这个函数:

... and so on (very messy!!!)I want to create a new column called is_liquid.My idea is that if the quantity string contains an l or L the is_liquid field in this row should get a 1 and if not 0.Here is what I've tried:I wrote this function:

def is_liquid(x):
    if x.str.contains('l'):
        return 1
    elif x.str.contains('L'):
        return 1
    else: return 0

(顺便说一句:如果某种东西以盎司"衡量，它是液态的吗?)

(BTW: if something is measured in 'oz' is it liquid?)

然后尝试应用它

df['is_liquid'] = df['quantity'].apply(is_liquid)

但是我得到的只是这个错误:

But all I get is this error:

AttributeError: 'str' object has no attribute 'str'

有人可以帮我吗?

推荐答案

使用 str.contains 和case=False表示布尔掩码，并通过 Series.astype :

Use str.contains with case=False for boolean mask and convert it to integers by Series.astype:

df['is_liquid']= df['liquids'].str.contains('L', case=False).astype(int)
print(df)
          liquids  is_liquid
0  365 g (314 ml)          1
1           992 g          0
2         2.46 kg          0
3      0,33 litre          1
4         15.87oz          0
5         250 ml           1
6             1 L          1
7           33 cl          1

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