如何根据Pandas中某列的值从DataFrame中选择行?
在SQL中,我会使用:
SELECT *
FROM table
WHERE column_name = some_value
如何根据Pandas中某列的值从DataFrame中选择行?
在SQL中,我会使用:
SELECT *
FROM table
WHERE column_name = some_value
当前回答
使用带有panda>=0.25.00的.query更灵活:
由于panda>=0.25.00,我们可以使用查询方法来使用panda方法过滤数据帧,甚至可以使用带有空格的列名。通常,列名中的空格会给出一个错误,但现在我们可以使用backtick(`)来解决这个问题-请参见GitHub:
# Example dataframe
df = pd.DataFrame({'Sender email':['ex@example.com', "reply@shop.com", "buy@shop.com"]})
Sender email
0 ex@example.com
1 reply@shop.com
2 buy@shop.com
将.query与方法str.endswith一起使用:
df.query('`Sender email`.str.endswith("@shop.com")')
输出
Sender email
1 reply@shop.com
2 buy@shop.com
此外,我们还可以通过在查询中用@前缀来使用局部变量:
domain = 'shop.com'
df.query('`Sender email`.str.endswith(@domain)')
输出
Sender email
1 reply@shop.com
2 buy@shop.com
其他回答
下面是一个简单的例子
from pandas import DataFrame
# Create data set
d = {'Revenue':[100,111,222],
'Cost':[333,444,555]}
df = DataFrame(d)
# mask = Return True when the value in column "Revenue" is equal to 111
mask = df['Revenue'] == 111
print mask
# Result:
# 0 False
# 1 True
# 2 False
# Name: Revenue, dtype: bool
# Select * FROM df WHERE Revenue = 111
df[mask]
# Result:
# Cost Revenue
# 1 444 111
tl;博士
熊猫相当于
select * from table where column_name = some_value
is
table[table.column_name == some_value]
多种条件:
table[(table.column_name == some_value) | (table.column_name2 == some_value2)]
or
table.query('column_name == some_value | column_name2 == some_value2')
代码示例
import pandas as pd
# Create data set
d = {'foo':[100, 111, 222],
'bar':[333, 444, 555]}
df = pd.DataFrame(d)
# Full dataframe:
df
# Shows:
# bar foo
# 0 333 100
# 1 444 111
# 2 555 222
# Output only the row(s) in df where foo is 222:
df[df.foo == 222]
# Shows:
# bar foo
# 2 555 222
在上面的代码中,是df[df.foo==222]行根据列值给出行,在本例中为222。
也可能出现多种情况:
df[(df.foo == 222) | (df.bar == 444)]
# bar foo
# 1 444 111
# 2 555 222
但在这一点上,我建议使用查询函数,因为它不那么冗长,并产生相同的结果:
df.query('foo == 222 | bar == 444')
要添加:您还可以执行df.groupby('column_name').get_group('column_desired_value').reset_index()以生成具有特定值的指定列的新数据帧。例如。,
import pandas as pd
df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(),
'B': 'one one two three two two one three'.split()})
print("Original dataframe:")
print(df)
b_is_two_dataframe = pd.DataFrame(df.groupby('B').get_group('two').reset_index()).drop('index', axis = 1)
#NOTE: the final drop is to remove the extra index column returned by groupby object
print('Sub dataframe where B is two:')
print(b_is_two_dataframe)
运行此命令可以:
Original dataframe:
A B
0 foo one
1 bar one
2 foo two
3 bar three
4 foo two
5 bar two
6 foo one
7 foo three
Sub dataframe where B is two:
A B
0 foo two
1 foo two
2 bar two
使用numpy.where可以获得更快的结果。
例如,使用unubtu的设置-
In [76]: df.iloc[np.where(df.A.values=='foo')]
Out[76]:
A B C D
0 foo one 0 0
2 foo two 2 4
4 foo two 4 8
6 foo one 6 12
7 foo three 7 14
时间比较:
In [68]: %timeit df.iloc[np.where(df.A.values=='foo')] # fastest
1000 loops, best of 3: 380 µs per loop
In [69]: %timeit df.loc[df['A'] == 'foo']
1000 loops, best of 3: 745 µs per loop
In [71]: %timeit df.loc[df['A'].isin(['foo'])]
1000 loops, best of 3: 562 µs per loop
In [72]: %timeit df[df.A=='foo']
1000 loops, best of 3: 796 µs per loop
In [74]: %timeit df.query('(A=="foo")') # slowest
1000 loops, best of 3: 1.71 ms per loop
在Pandas的更新版本中,受文档启发(查看数据):
df[df["colume_name"] == some_value] #Scalar, True/False..
df[df["colume_name"] == "some_value"] #String
通过将子句放在括号()中,并用&和|(和/或)组合来组合多个条件。这样地:
df[(df["colume_name"] == "some_value1") & (pd[pd["colume_name"] == "some_value2"])]
其他过滤器
pandas.notna(df["colume_name"]) == True # Not NaN
df['colume_name'].str.contains("text") # Search for "text"
df['colume_name'].str.lower().str.contains("text") # Search for "text", after converting to lowercase