1、使用loc[]实现
1)要选择列值等于some_value
df.loc[df['column_name'] == some_value]
2)要选择列包含在可迭代的some_values
df.loc[df['column_name'].isin(some_values)]
3)多个条件可以使用&
df.loc[(df['column_name'] >= A) & (df['column_name'] <= B)]
4)要选择列值不等于some_value
df.loc[df['column_name'] != some_value]
5)isin返回布尔值是在其中的条件,不在其中可以使用~
df.loc[~df['column_name'].isin(some_values)]
例如,
import pandas as pd import numpy as np df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(), 'B': 'one one two three two two one three'.split(), 'C': np.arange(8), 'D': np.arange(8) * 2}) print(df) # A B C D # 0 foo one 0 0 # 1 bar one 1 2 # 2 foo two 2 4 # 3 bar three 3 6 # 4 foo two 4 8 # 5 bar two 5 10 # 6 foo one 6 12 # 7 foo three 7 14 print(df.loc[df['A'] == 'foo']) # 输出: # 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 print(df.loc[df['B'].isin(['one','three'])]) # 输出: # A B C D # 0 foo one 0 0 # 1 bar one 1 2 # 3 bar three 3 6 # 6 foo one 6 12 # 7 foo three 7 14 df = df.set_index(['B']) print(df.loc['one']) # 输出: # A C D # B # one foo 0 0 # one bar 1 2 # one foo 6 12 df.loc[df.index.isin(['one','two'])] # 输出: # A C D # B # one foo 0 0 # one bar 1 2 # two foo 2 4 # two foo 4 8 # two bar 5 10 # one foo 6 12
2、使用query()
pandas >= 0.25.0 可以使用该query()
方法来过滤带有pandas 方法的DataFrame。通常,列名中的空格会产生错误,但现在我们可以使用反勾号('
)来解决这个问题。
例如,
import pandas as pd import numpy as np df = pd.DataFrame({'Sender email':['ex@example.com', "reply@cjavapy.com", "buy@cjavapy.com"]}) # Sender email # 0 ex@example.com # 1 reply@cjavapy.com # 2 buy@cjavapy.com df.query('`Sender email`.str.endswith("@cjavapy.com")') # Sender email # 1 reply@cjavapy.com # 2 buy@cjavapy.com domain = 'cjavapy.com' df.query('`Sender email`.str.endswith(@domain)') # Sender email # 1 reply@cjavapy.com # 2 buy@cjavapy.com