选择列值column_name等于标量some_value的行时，我们使用==:
df.loc[df['column_name'] == some_value]
或使用.query()
df.query('column_name == some_value')
在一个具体的例子中：
import pandas as pd
import numpy as np
df = pd.DataFrame({'Col1': 'what are men to rocks and mountains'.split(),
'Col2': 'the curves of your lips rewrite history.'.split(),
'Col3': np.arange(7),
'Col4': np.arange(7) * 8})
print(df)
Col1 Col2 Col3 Col4
0 what the 0 0
1 are curves 1 8
2 men of 2 16
3 to your 3 24
4 rocks lips 4 32
5 and rewrite 5 40
6 mountains history 6 48
查询可以是
rocks_row = df.loc[df['Col1'] == "rocks"]
输出
print(rocks_row)
Col1 Col2 Col3 Col4
4 rocks lips 4 32
我想通过一个值列表来查询数据帧，该数据帧输出一个＆＃34;正确查询列表＆＃34;。
要执行的查询将在列表中，例如
list_match = ['men', 'curves', 'history']
将输出满足此条件的所有行，即
matches = pd.concat([df1, df2, df3])
其中
df1 = df.loc[df['Col1'] == "men"]
df2 = df.loc[df['Col1'] == "curves"]
df3 = df.loc[df['Col1'] == "history"]
我的想法是创建一个接收
的函数output = []
def find_queries(dataframe, column, value, output):
for scalar in value:
query = dataframe.loc[dataframe[column] == scalar]]
output.append(query) # append all query results to a list
return pd.concat(output) # return concatenated list of dataframes
然而，这看起来特别慢，并没有实际利用熊猫数据结构。什么是＆＃34;标准＆＃34;通过pandas数据框传递查询列表的方法？
编辑：这是如何转化为更复杂的＆＃34;在熊猫中查询？例如where有HDF5文件吗？
df.to_hdf('test.h5','df',mode='w',format='table',data_columns=['A','B'])
pd.read_hdf('test.h5','df')
pd.read_hdf('test.h5','df',where='A=["foo","bar"] & B=1')
答案 0 :(得分：1)
处理此问题的最佳方法是使用布尔系列索引行，就像在R中一样。
以你的df为例，
In [5]: df.Col1 == "what"
Out[5]:
0 True
1 False
2 False
3 False
4 False
5 False
6 False
Name: Col1, dtype: bool
In [6]: df[df.Col1 == "what"]
Out[6]:
Col1 Col2 Col3 Col4
0 what the 0 0
现在我们将它与pandas isin功能结合起来。
In [8]: df[df.Col1.isin(["men","rocks","mountains"])]
Out[8]:
Col1 Col2 Col3 Col4
2 men of 2 16
4 rocks lips 4 32
6 mountains history. 6 48
要对多列进行过滤，我们可以将它们与＆amp;链接在一起和|像这样的运营商。
In [10]: df[df.Col1.isin(["men","rocks","mountains"]) | df.Col2.isin(["lips","your"])]
Out[10]:
Col1 Col2 Col3 Col4
2 men of 2 16
3 to your 3 24
4 rocks lips 4 32
6 mountains history. 6 48
In [11]: df[df.Col1.isin(["men","rocks","mountains"]) & df.Col2.isin(["lips","your"])]
Out[11]:
Col1 Col2 Col3 Col4
4 rocks lips 4 32
答案 1 :(得分：1)
如果我正确理解了您的问题，您可以使用布尔索引作为@uhjish has already shown in his answer或使用query()方法：
$ line=$(sed '
s/"/"Dcom.sun.management.jmxremote=true -Djava.rmi.server.hostname=x.x.x.x /
s/authenticate/&=false/
' <<<"$line")
$ echo "$line"
KAFKA_JMX_OPTS="Dcom.sun.management.jmxremote=true -Djava.rmi.server.hostname=x.x.x.x -Dcom.sun.management.jmxremote.authenticate=false -Dcom.sun.management.jmxremote.ssl=false "
In [30]: search_list = ['rocks','mountains']
In [31]: df
Out[31]:
Col1 Col2 Col3 Col4
0 what the 0 0
1 are curves 1 8
2 men of 2 16
3 to your 3 24
4 rocks lips 4 32
5 and rewrite 5 40
6 mountains history. 6 48
方法：
.query()使用布尔索引：
In [32]: df.query('Col1 in @search_list and Col4 > 40')
Out[32]:
Col1 Col2 Col3 Col4
6 mountains history. 6 48
In [33]: df.query('Col1 in @search_list')
Out[33]:
Col1 Col2 Col3 Col4
4 rocks lips 4 32
6 mountains history. 6 48
更新：使用功能：
In [34]: df.ix[df.Col1.isin(search_list) & (df.Col4 > 40)]
Out[34]:
Col1 Col2 Col3 Col4
6 mountains history. 6 48
In [35]: df.ix[df.Col1.isin(search_list)]
Out[35]:
Col1 Col2 Col3 Col4
4 rocks lips 4 32
6 mountains history. 6 48
包括调试信息（打印查询）：
def find_queries(df, qry, debug=0, **parms):
if debug:
print('[DEBUG]: Query:\t' + qry.format(**parms))
return df.query(qry.format(**parms))
In [31]: find_queries(df, 'Col1 in {Col1} and Col4 > {Col4}', Col1='@search_list', Col4=40)
...:
Out[31]:
Col1 Col2 Col3 Col4
6 mountains history. 6 48
In [32]: find_queries(df, 'Col1 in {Col1} and Col4 > {Col4}', Col1='@search_list', Col4=10)
Out[32]:
Col1 Col2 Col3 Col4
4 rocks lips 4 32
6 mountains history. 6 48