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Python 从数据帧返回单个单元格值_Python_Pandas - Fatal编程技术网

Python 从数据帧返回单个单元格值

Python 从数据帧返回单个单元格值,python,pandas,Python,Pandas,我想问一个问题,这是对这个主题的扩展: 此线程中的代码如下所示: import pandas as pd import numpy as np df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(), 'B': 'one one two three two two one three'.split(), 'C': np.arange(8), 'D': np.a

我想问一个问题,这是对这个主题的扩展:

此线程中的代码如下所示:

import pandas as pd
import numpy as np
df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(),
               'B': 'one one two three two two one three'.split(),
               'C': np.arange(8), 'D': np.arange(8) * 2})
print(df)
#      A      B  C   D
# 0  foo    one  0   0
# 1  bar    one  1   2
# 2  foo    two  2   4
# 3  bar  three  3   6
# 4  foo    two  4   8
# 5  bar    two  5  10
# 6  foo    one  6  12
# 7  foo  three  7  14

print(df.loc[df['D'] == 14])
这将产生以下结果:

   A    B      C   D
7  foo  three  7  14

基于上面的代码,我如何返回一个“值”而不是一行。也就是说,如何返回与整行相反的值
'7'
或值
'foo'

import pandas as pd
import numpy as np
df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(),
               'B': 'one one two three two two one three'.split(),
               'C': np.arange(8), 'D': np.arange(8) * 2})

print(df.loc[df['D'] == 14]['A'].index.values)

>>>[7]

print(df.loc[df['D'] == 14]['A'].values)

>>>['foo']

df.loc[df['D']==14][colName]或df.loc[df['D']==14]。值[index]非常棒。非常感谢你们两位。这些返回长度为1的nArray,而不是实际值。我的解决方案是min(df['A'][df['D']==14]),这种语法对我来说更符合逻辑,尽管使用min()似乎是一种错误。当然还有更好的方法?@smartse-我通常会使用索引(
[0]
)来获取第一个值,而不是使用
min()
,如下所示:
df[df['D']==14].loc[:,'a'].values[0]
。尽管如此,我只希望返回一个值。