根据字母"l"或"L"是否在另一列的字符串中创建新列 [英] creating new column based on whether the letter 'l' or 'L' is in the string of another column

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

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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