Question

我想根据同一行中某个列中的值获取另一列的值。

示例：

对于商家ID =＆＃39; 123＆＃39;，我想检索business_name

DF：

biz_id  biz_name
123      chew
456      bite
123      chew

代码：

df['biz_name'].loc[df['biz_id'] == 123]

给我回复：

chew
chew

如何以字符串格式获得'chew'的1个值？

Answer 1

使用idxmax获取第一个最大值的索引

df.loc[df.biz_id.eq(123).idxmax(), 'biz_name']

'chew'

Answer 2

您可以使用iloc或iat来选择Series的第一个值：

print (df.loc[df['biz_id'] == 123, 'biz_name'].iloc[0])
chew

或者：

print (df.loc[df['biz_id'] == 123, 'biz_name'].iat[0])
chew

使用query：

print (df.query('biz_id == 123')['biz_name'].iloc[0])
chew

或者在list或numpy array中选择第一个值：

print (df.loc[df['biz_id'] == 123, 'biz_name'].tolist()[0])
chew

print (df.loc[df['biz_id'] == 123, 'biz_name'].values[0])
chew

<强>计时：

In [18]: %timeit (df.loc[df['biz_id'] == 123, 'biz_name'].iloc[0])
1000 loops, best of 3: 399 µs per loop

In [19]: %timeit (df.loc[df['biz_id'] == 123, 'biz_name'].iat[0])
The slowest run took 4.16 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 391 µs per loop

In [20]: %timeit (df.query('biz_id == 123')['biz_name'].iloc[0])
The slowest run took 4.39 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 1.75 ms per loop

In [21]: %timeit (df.loc[df['biz_id'] == 123, 'biz_name'].tolist()[0])
The slowest run took 4.18 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 384 µs per loop

In [22]: %timeit (df.loc[df['biz_id'] == 123, 'biz_name'].values[0])
The slowest run took 5.32 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 370 µs per loop

In [23]: %timeit (df.loc[df.biz_id.eq(123).idxmax(), 'biz_name'])
1000 loops, best of 3: 517 µs per loop

pandas df locate只保留第一项

2 个答案: