Question

这是方程式。例子

Numbers     | Date      | Mean
1000        |12/1/2018  | 1
1002        |12/2/2018  | 0
1003        |12/3/2018  | 0.5
 0          |12/4/2018  | 0.6
 0          |12/5/2018  | 0.4
 0          |12/6/2018  |0.1
 0          |12/7/2018  | -0.7
 0          |12/8/2018  | 0.2
 0          |12/9/2018  | -0.1

这就是我想要的

|Numbers | Date      | Mean | Multiplication |
| ------ |-----------|------|----------------|
|1000    | 12/1/2018 | 1    | 1000           |
|1002    | 12/2/2018 | 0    | 0*1000= 0      |
|1003    | 12/3/2018 | 0.5  | 0.5*1002=501   |
|0       | 12/4/2018 | 0.6  | 1003*0.6=601.8 |
|0       | 12/5/208  | 0.4  | 601.8*0.4
|0       | 12/6/2018 | 0.1  | 601.8*0.4*0.1  |
|0       |12/7/2018  | -0.7 |601.8*0.4*0.1*-0.7| 
 0       |12/8/2018  | 0.2  |601.8*0.4*0.1*-0.7*0.2
 0       |12/9/2018  | -0.1 |601.8*0.4*0.1*-0.7*0.2*-0.1

数据已经在数据框中，我正在使用熊猫函数

Answer 1

如果使用Pandas数据框，则可以使用shift()方法：

df['Multiplication'] = df.Mean * df.Numbers.shift(1)
df.loc[0, 'Multiplication'] = df.Numbers[0]
for i in range(len(df[df.Numbers.shift(1) == 0])):
    df.loc[df[df.Numbers.shift(1) == 0].index, 'Multiplication'] = df[df.Numbers.shift(1) == 0].Mean.values * df[df.index.isin(df[df.Numbers.shift(1) == 0].index-1)].Multiplication.values

具有以下输出：

第一行没有两个数字可以相乘，因此更新值是在相乘之后得出的。

现在满足零数字值的要求。

像@ Raj006所建议的那样，破坏一点代码：

# return just the rows that match the condition (Numbers column 0 for the row before)
df[df.Numbers.shift(1) == 0].index

# update the values for the column Multiplication with the matching rows
df.loc[df[df.Numbers.shift(1) == 0].index, 'Multiplication']

# the value to be update is the [Mean value for the matching rows] * [rows before the matching Multiplication value]
df[df.Numbers.shift(1) == 0].Mean.values * df[df.index.isin(df[df.Numbers.shift(1) == 0].index-1)].Multiplication.values

编辑：不幸的是，一旦计算取决于之前的计算，我必须使用for循环来运行与匹配行一样多的次数。

Answer 2

@Daniel Labbe回答了最初的要求，这是正确的。向他+1以获取shift（）方法。然后，用户需求发生了变化。因此，这是我对最新要求的回答。

#import pandas for managing data with dataframe
import pandas as pd
#import tabulate to print your data frame as table
from tabulate import tabulate
#Create a data dictionary
myData={'Numbers':[1000,1002,1003,0,0,0,0,0,0],'Date':['12/1/2018','12/2/2018','12/3/2018','12/4/2018','12/5/2018','12/6/2018','12/7/2018','12/8/2018','12/9/2018'],'Mean':[1,0,0.5,0.6,0.4,0.1,-0.7,0.2,-0.1]}
#Create a data frame from the data dictionary using pandas. User mentioned that the data is already in the
#pandas data frame
myDataFrame=pd.DataFrame(myData)
#Print your final table (just pretty print)
print(tabulate(myDataFrame, headers='keys', tablefmt='psql'))
#Declare a list
MultiplicationList=[]
#Declare a constant
StorePreviousValue=0
for i in range(0,len(myDataFrame['Numbers'])):
    #If it is the first row then use the Number
    if i==0:
        #Append the value to the list
        MultiplicationList.append(myDataFrame['Numbers'][i])
    else:
        #If it is not the first row, and the value in the first column of the previous row is '0'
        #multiply Mean with the previous multiplication result
        if myDataFrame['Numbers'][i-1]==0:
            StorePreviousValue=StorePreviousValue*myDataFrame['Mean'][i]
        #If it is not the first row, and the value in the first column of the previous row is not '0'
        #(should probably say greate than '0', but the question is not clear about that), then 
        #multiply Mean with the Number in the first column of the previous row
        else:
            StorePreviousValue=myDataFrame['Numbers'][i-1]*myDataFrame['Mean'][i]
        #Append the value to the list
        MultiplicationList.append(StorePreviousValue)
#Create a new column in the data frame and pass the list as the value
myDataFrame['Multiplication']=MultiplicationList
#Print your final table (just pretty print)
print(tabulate(myDataFrame, headers='keys', tablefmt='psql'))

这是输出

+----+-----------+-----------+--------+
|    |   Numbers | Date      |   Mean |
|----+-----------+-----------+--------|
|  0 |      1000 | 12/1/2018 |    1   |
|  1 |      1002 | 12/2/2018 |    0   |
|  2 |      1003 | 12/3/2018 |    0.5 |
|  3 |         0 | 12/4/2018 |    0.6 |
|  4 |         0 | 12/5/2018 |    0.4 |
|  5 |         0 | 12/6/2018 |    0.1 |
|  6 |         0 | 12/7/2018 |   -0.7 |
|  7 |         0 | 12/8/2018 |    0.2 |
|  8 |         0 | 12/9/2018 |   -0.1 |
+----+-----------+-----------+--------+
+----+-----------+-----------+--------+------------------+
|    |   Numbers | Date      |   Mean |   Multiplication |
|----+-----------+-----------+--------+------------------|
|  0 |      1000 | 12/1/2018 |    1   |      1000        |
|  1 |      1002 | 12/2/2018 |    0   |         0        |
|  2 |      1003 | 12/3/2018 |    0.5 |       501        |
|  3 |         0 | 12/4/2018 |    0.6 |       601.8      |
|  4 |         0 | 12/5/2018 |    0.4 |       240.72     |
|  5 |         0 | 12/6/2018 |    0.1 |        24.072    |
|  6 |         0 | 12/7/2018 |   -0.7 |       -16.8504   |
|  7 |         0 | 12/8/2018 |    0.2 |        -3.37008  |
|  8 |         0 | 12/9/2018 |   -0.1 |         0.337008 |
+----+-----------+-----------+--------+------------------+

如果您没有熊猫或制表，请使用点安装熊猫 pip install制表

如果您不熟悉点子，请使用google。该答案假设您知道如何从文件中读取数据并创建数据字典。如果您不这样做，那将是另一个要问的问题。

如何在python中不同行的列之间相乘

2 个答案: