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')

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

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'])]]

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