Python 如何正确获取pandas中的单个单元格：loc[index，column]VS get_value（index，column）

使用哪种方法（在性能和可靠性方面）更好地从pandas

数据帧获取单个单元格：get_value（）或loc[]？
您最终可以在以下内容中找到信息：

用于显式获取值（相当于不推荐的df.get_值（'a'，'a'））

但如果使用它，则没有警告

但如果检查：

从一维数组快速查找值仅当您知道自己在做什么时才使用此选项

所以我认为更好的是使用

计时：

df = pd.DataFrame({'A':[1,2,3],
                   'B':[4,5,6],
                   'C':[7,8,9],
                   'D':[1,3,5],
                   'E':[5,3,6],
                   'F':[7,4,3]})

print (df)

In [93]: %timeit (df.loc[0, 'A'])
The slowest run took 6.40 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 177 µs per loop

In [96]: %timeit (df.at[0, 'A'])
The slowest run took 17.01 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 7.61 µs per loop

In [94]: %timeit (df.get_value(0, 'A'))
The slowest run took 23.49 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 3.36 µs per loop

get\u value（）
-将更快更合适我不理解文档中关于
弃用get\u value（）
的这句话。。。熊猫本身在内部使用
get_value（）
查看iat/iloc、at/loc和ix之间的差异请参见@JulienMarrec-谢谢。我还发现
df = pd.DataFrame({'A':[1,2,3],
                   'B':[4,5,6],
                   'C':[7,8,9],
                   'D':[1,3,5],
                   'E':[5,3,6],
                   'F':[7,4,3]})

print (df)

In [93]: %timeit (df.loc[0, 'A'])
The slowest run took 6.40 times longer than the fastest. This could mean that an intermediate result is being cached.
10000 loops, best of 3: 177 µs per loop

In [96]: %timeit (df.at[0, 'A'])
The slowest run took 17.01 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 7.61 µs per loop

In [94]: %timeit (df.get_value(0, 'A'))
The slowest run took 23.49 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 3.36 µs per loop