python2.7：更改dataframe的列值的差异 [英] python2.7: change difference for a column value of dataframe

问题描述

我有一个如下所示的数据帧（df）（仅作为示例），也许有10个或更多数据帧：

 日期ab 
 0 2010-01-01 12 15 
 1 2010-01-02 13 20 
 2 2010-01-03 14 23 
 3 2010-01-04 15 24 
 4 2010-01-05 16 25 
 5 2010-01-08 17 15 
 6 2010-01-09 180160 
 ..... ...................... 
 1000 2013-01-05 310320 
  但有一个例外，当日期为'2010-01-09'（仅作为示例），并计算b'2010-01-09'的变化百分比时，b在'2010-  01-08应为10倍，仅此一次，其他日期应使用原始值，我的意思是没有10倍。通常，我通过以下代码计算更改百分比：
  df ['b_diff'] = df2 ['b'] /（df2 ['b']。shift（）-1 
  
但是当日期为：' 2010-01-09'。
我认为代码应为：
  df ['b_diff'] = df2 ['b'] / 10 *（df2 ['b']。shift（））-1 
  
您能告诉我如何处理此问题吗？
 
 
 
谢谢！
解决方案
您可以使用  pct_change  ，但首先将 b 的值除以条件：
  dates = ['2010-01-09'，'2011-01-09'] 
m = df2 ['date']。isin（dates）
 df2 .loc [m，'b'] = df2 ['b'] / 10 
 
 df2 ['b_diff'] = df2 ['b']。pct_change（）
打印（df2 ）
日期ab b_diff 
 0 2010-01-01 12 15.0 NaN 
 1 2010-01-02 13 20。 0 0.333333 
 2 2010-01-03 14 23.0 0.150000 
 3 2010-01-04 15 24.0 0.043478 
 4 2010-01-05 16 25.0 0.041667 
 5 2010-01- 08 17 15.0 -0.400000 
 6 2010-01-09 180 16.0 0.066667 
  
替代解决方案： 
  dates = ['2010-01-09'，'2011-01-09'] 
m = df2 [ 'date']。isin（dates）
 
 df2 ['b'] = df2 ['b']。mask（m，df2 ['b'] / 10）
 df2 [ 'b_diff'] = df2 ['b']。pct_change（）
 print（df2）
 date ab b_diff 
 0 2010-01-01 12 15.0 NaN 
 1 2010- 01-02 13 20.0 0.333333 
 2 2010-01-03 14 23.0 0.150000 
 3 2010-01-04 15 24.0 0.043478 
 4 2010-01-05 16 25.0 0.041667 
 5 2010-01-08 17 15.0 -0.400000 
 6 2010-01-09 180 16.0 0.066667 
  
 
I have a dataframe(df) like as following(just example), there are maybe 10 or more dataframes:
     date              a       b
  0     2010-01-01     12      15
  1     2010-01-02     13      20
  2     2010-01-03     14      23
  3     2010-01-04     15      24
  4     2010-01-05     16      25
  5     2010-01-08     17      15
  6     2010-01-09     180     160
  ................................
  1000     2013-01-05     310     320
I want to calculate the change percentage of b column value in the dataframe.
But there is a exception that when the date is '2010-01-09' (just a example), and calculate the change percentage of b '2010-01-09' , the value of b in'2010-01-08' should be 10 times, just this time, other dates should use the original value, I mean no 10 times. In generally, I calculate the change percent by the following code:
df['b_diff'] = df2['b']/(df2['b'].shift() -1
But when the date is: '2010-01-09'.
    I think the code should be:
 df['b_diff'] = df2['b']/10*(df2['b'].shift()) -1 
Could you tell me how to process with this issue?

Thanks!
解决方案
You can use pct_change, but first divide value of b by condition:
dates = ['2010-01-09','2011-01-09']
m = df2['date'].isin(dates)
df2.loc[m, 'b'] =  df2['b'] / 10

df2['b_diff'] = df2['b'].pct_change()
print (df2)
        date    a     b    b_diff
0 2010-01-01   12  15.0       NaN
1 2010-01-02   13  20.0  0.333333
2 2010-01-03   14  23.0  0.150000
3 2010-01-04   15  24.0  0.043478
4 2010-01-05   16  25.0  0.041667
5 2010-01-08   17  15.0 -0.400000
6 2010-01-09  180  16.0  0.066667
Alternative solution:
dates = ['2010-01-09','2011-01-09']
m = df2['date'].isin(dates)

df2['b'] = df2['b'].mask(m, df2['b'] / 10)
df2['b_diff'] = df2['b'].pct_change()
print (df2)
        date    a     b    b_diff
0 2010-01-01   12  15.0       NaN
1 2010-01-02   13  20.0  0.333333
2 2010-01-03   14  23.0  0.150000
3 2010-01-04   15  24.0  0.043478
4 2010-01-05   16  25.0  0.041667
5 2010-01-08   17  15.0 -0.400000
6 2010-01-09  180  16.0  0.066667


                        
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