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

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

对于Pandas中给定值的多个列中仅选择特定列：

select col_name1, col_name2 from table where column_name = some_value.

选项位置：

df.loc[df['column_name'] == some_value, [col_name1, col_name2]]

或查询：

df.query('column_name == some_value')[[col_name1, col_name2]]

如果您想重复查询数据帧，并且速度对您很重要，最好的方法是将数据帧转换为字典，然后通过这样做，您可以将查询速度提高数千倍。

my_df = df.set_index(column_name)
my_dict = my_df.to_dict('index')

制作my_dict字典后，您可以浏览：

if some_value in my_dict.keys():
   my_result = my_dict[some_value]

如果column_name中有重复值，则无法创建字典。但您可以使用：

my_result = my_df.loc[some_value]

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

您也可以使用.apply：

df.apply(lambda row: row[df['B'].isin(['one','three'])])

它实际上按行工作（即，将函数应用于每一行）。

输出为

   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

结果与@unsubu提到的使用相同

df[[df['B'].isin(['one','three'])]]