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
我正在使用非常混乱的 Open Food Facts 数据集.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(非常凌乱!!!)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
,则该行中的 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:如果某物以盎司"来衡量,它是液体吗?)
(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
with case=False
用于布尔掩码并将其转换为 integer
sSeries.astype
一个>:
Use str.contains
with case=False
for boolean mask and convert it to integer
s 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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