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