当选择列值column_name等于标量some_value的行时,我们使用==:
df.loc[df['column_name'] == some_value]
或者使用.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"]
哪个输出
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“正确的查询”列表.
要执行的查询将在列表中,例如
list_match = ['men', 'curves', 'history']
这将输出满足这种条件的所有行,即
matches = pd.concat([df1, df2, df3])
哪里
df1 = df.loc[df['Col1'] == "men"]
df2 = df.loc[df['Col1'] == "curves"]
df3 = df.loc[df['Col1'] == "history"]
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
但是,这看起来特别慢,实际上并没有利用pandas数据结构.通过pandas数据框传递查询列表的“标准”方法是什么?
编辑:这如何转化为熊猫中“更复杂”的查询?例如哪里有HDF5文件?
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')
解决方法:
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:
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
.query()方法:
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
使用布尔索引:
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
更新:使用功能:
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
In [40]: find_queries(df, 'Col1 in {Col1} and Col4 > {Col4}', Col1='@search_list', Col4=10, debug=1)
[DEBUG]: Query: Col1 in @search_list and Col4 > 10
Out[40]:
Col1 Col2 Col3 Col4
4 rocks lips 4 32
6 mountains history. 6 48
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