ICode9

精准搜索请尝试: 精确搜索
首页 > 编程语言> 文章详细

python-编写函数以计算熊猫中按行元素的最佳方法是什么?

2019-11-09 01:56:50  阅读:16  来源: 互联网

标签:pandas dataframe python function



我有一个基本表,如:

enter image description here

col1是一列独立的值,col2是基于“国家和类型”组合的汇总.我想使用以下逻辑来计算col3到col5列:

> col3:col1中的元素与col1的总数之比
> col4:col1中的元素与col2中相应元素的比率
> col5:col3和col4中按行元素乘积的自然指数

我写了下面的函数来实现这个目的:

def calculate(df):
  for i in range(len(df)):
    df['col3'].loc[i] = df['col1'].loc[i]/sum(df['col1'])
    df['col4'].loc[i] = df['col1'].loc[i]/df['col2'].loc[i]
    df['col5'].loc[i] = np.exp(df['col3'].loc[i]*df['col4'].loc[i])
  return df

该函数会执行,并给我预期的结果,但是笔记本还会引发警告:

SettingWithCopyWarning:

A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation:
07001

我不确定是否在这里编写最佳功能.任何帮助,将不胜感激!谢谢.

解决方法:

我认为最好避免在熊猫中应用和循环,因此更好,更快地使用虚拟化解决方案:

df = pd.DataFrame({'col1':[4,5,4,5,5,4],
                   'col2':[7,8,9,4,2,3],
                   'col3':[1,3,5,7,1,0],
                   'col4':[5,3,6,9,2,4],
                   'col5':[1,4,3,4,0,4]})

print (df)
   col1  col2  col3  col4  col5
0     4     7     1     5     1
1     5     8     3     3     4
2     4     9     5     6     3
3     5     4     7     9     4
4     5     2     1     2     0
5     4     3     0     4     4

df['col3'] = df['col1']/(df['col1']).sum()
df['col4'] = df['col1']/df['col2']
df['col5'] = np.exp(df['col3']*df['col4'])
print (df)
   col1  col2      col3      col4      col5
0     4     7  0.148148  0.571429  1.088343
1     5     8  0.185185  0.625000  1.122705
2     4     9  0.148148  0.444444  1.068060
3     5     4  0.185185  1.250000  1.260466
4     5     2  0.185185  2.500000  1.588774
5     4     3  0.148148  1.333333  1.218391

时间:

df = pd.DataFrame({'col1':[4,5,4,5,5,4],
                   'col2':[7,8,9,4,2,3],
                   'col3':[1,3,5,7,1,0],
                   'col4':[5,3,6,9,2,4],
                   'col5':[1,4,3,4,0,4]})

#print (df)

#6000 rows
df = pd.concat([df] * 1000, ignore_index=True)

In [211]: %%timeit
     ...: df['col3'] = df['col1']/(df['col1']).sum()
     ...: df['col4'] = df['col1']/df['col2']
     ...: df['col5'] = np.exp(df['col3']*df['col4'])
     ...: 
1.49 ms ± 104 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

不幸的是,对于此示例,循环解决方案的速度确实很慢,因此仅在60行DataFrame中进行了测试:

#60 rows
df = pd.concat([df] * 10, ignore_index=True)

In [3]: %%timeit
   ...: (calculate(df))
   ...: 
C:\Anaconda3\lib\site-packages\pandas\core\indexing.py:194: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  self._setitem_with_indexer(indexer, value)
10.2 s ± 410 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)


标签:pandas,dataframe,python,function

专注分享技术,共同学习,共同进步。侵权联系[admin#icode9.com]

Copyright (C)ICode9.com, All Rights Reserved.

ICode9版权所有