如何根据Pandas中某列的值从DataFrame中选择行?

在SQL中,我会使用:

SELECT *
FROM table
WHERE column_name = some_value

当前回答

很好的答案。只有当数据帧的大小接近百万行时,许多方法在使用df[df['col']==val]时往往需要很长时间。我希望“another_column”的所有可能值都对应于“some_column“中的特定值(在本例中是在字典中)。这起作用很快。

s=datetime.datetime.now()

my_dict={}

for i, my_key in enumerate(df['some_column'].values): 
    if i%100==0:
        print(i)  # to see the progress
    if my_key not in my_dict.keys():
        my_dict[my_key]={}
        my_dict[my_key]['values']=[df.iloc[i]['another_column']]
    else:
        my_dict[my_key]['values'].append(df.iloc[i]['another_column'])
        
e=datetime.datetime.now()

print('operation took '+str(e-s)+' seconds')```

其他回答

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]

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')

您可以在函数中使用loc(方括号):

# Series
s = pd.Series([1, 2, 3, 4]) 
s.loc[lambda x: x > 1]
# s[lambda x: x > 1]

输出:

1    2
2    3
3    4
dtype: int64

or

# DataFrame
df = pd.DataFrame({'A': [1, 2, 3], 'B': [10, 20, 30]})
df.loc[lambda x: x['A'] > 1]
# df[lambda x: x['A'] > 1]

输出:

   A   B
1  2  20
2  3  30

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

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