Python 数据框按列值过滤行
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Dataframe filtering rows by column values
提问by Seunghun Choi
I have a Dataframe df
我有一个数据框 df
Num1 Num2
one 1 0
two 3 2
three 5 4
four 7 6
five 9 8
I want to filter rows that have value bigger than 3 in Num1 and smaller than 8 in Num2.
我想过滤 Num1 中值大于 3 且 Num2 中值小于 8 的行。
I tried this
我试过这个
df = df[df['Num1'] > 3 and df['Num2'] < 8]
but the error occurred.
但发生了错误。
ValueError: The truth value of a Series is ambiguous.
ValueError：系列的真值不明确。
so I used
所以我用
df = df[df['Num1'] > 3]
df = df[df['Num2'] < 8]
I think the code can be shorter.
我认为代码可以更短。
Is there any other way?
有没有其他办法？
回答by jezrael
You need add ()because operator precedence with bit-wise operator &:
您需要添加，()因为运算符优先级为按位运算符&：
df1 = df[(df['Num1'] > 3) & (df['Num2'] < 8)]
print (df1)
Num1 Num2
three 5 4
four 7 6
Better explanation is here.
更好的解释是here。
Or if need shortest code use query:
或者如果需要最短的代码使用query：
df1 = df.query("Num1 > 3 and Num2 < 8")
print (df1)
Num1 Num2
three 5 4
four 7 6
df1 = df.query("Num1 > 3 & Num2 < 8")
print (df1)
Num1 Num2
three 5 4
four 7 6
回答by Willem Van Onsem
Yes, you can use the &operator:
是的，您可以使用&运算符：
df = df[(df['Num1'] > 3) & (df['Num2'] < 8)]
# ^ & operatorThis is because andworks on the truthiness valueof the two operands, whereas the &operator can be defined on arbitrary data structures.
这是因为and适用于两个操作数的真实值，而&运算符可以定义在任意数据结构上。
The brackets are mandatory here, because &binds shorter than >and <, so without brackets, Python would read the expression as df['Num1'] > (3 & df['Num2']) < 8.
括号在这里是强制性的，因为&绑定比>and短<，所以如果没有括号，Python 会将表达式读作df['Num1'] > (3 & df['Num2']) < 8.
Note that you can use the |operator as a logical or.
请注意，您可以将|运算符用作逻辑或。
