LOC 函数中的 Pandas 使用和运算符
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Pandas use and operator in LOC function
提问by jxn
i want to have 2 conditions in the locfunction but the &&or andoperators dont seem to work.:
我想在loc函数中有 2 个条件，但&&orand运算符似乎不起作用。：
df:
df：
business_id ratings review_text
xyz 2 'very bad'
xyz 1 'passable'
xyz 3 'okay'
abc 2 'so so'
mycode:
i am trying to gather all review_textwhose ratings are < 3and have id = xyzinto a list
mycode的：我试图收集所有review_text它的收视率< 3，并有id = xyz到一个列表
id = 'xyz'
mylist = df.loc[df['ratings'] < 3 and df[business_id] ==id,'review_text'].values.tolist()
i should get:
我应该得到：
['very bad','passable']
This code doesnt work and i get the error:
此代码不起作用，我收到错误消息：
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
how do i use the andoperator correctly here?
我如何and在这里正确使用运算符？
采纳答案by jezrael
You need &for andlogical operator, because need element-wise and, see boolean indexing:
你需要&的and逻辑运算符，因为需要逐元素and，见布尔索引：
id = 'xyz'
mylist=df.loc[(df['ratings'] < 3) & (df['business_id'] == id),'review_text'].values.tolist()
print (mylist)
['very bad', 'passable']
回答by piRSquared
Using query
使用 query
df.query('ratings < 3 & business_id == @id').review_text.tolist()
["'very bad'", "'passable'"]
回答by B. Kanani
use ~ for not logical operation
使用 ~ 进行非逻辑操作
For example:
例如：
train.loc[(train['MiscVal'] == 0) & (~train['MiscFeature'].isna())]
