Question

在简化的数据框中：

import pandas as pd 

df1 = pd.DataFrame({'350': [7.898167, 6.912074, 6.049002, 5.000357, 4.072320],
                '351': [8.094912, 7.090584, 6.221289, 5.154516, 4.211746],
                '352': [8.291657, 7.269095, 6.393576, 5.308674, 4.351173],
                '353': [8.421007, 7.374317, 6.496641, 5.403691, 4.439815],
                '354': [8.535562, 7.463452, 6.584512, 5.485725, 4.517310],
                '355': [8.650118, 7.552586, 6.672383, 4.517310, 4.594806]},
                 index=[1, 2, 3, 4, 5])

int_range = df1.columns.astype(float)
a = 0.005
b = 0.837

我想求解一个方程，如下图所示：

I 等于数据框中的值。 x是 int_range 值，因此在本例中为dx = 1，从350到355。 a 和 b 是可选常量

我需要获取一个数据框作为每每行

的输出

现在我做这样的事情，但是我不确定它是正确的：

dict_INT = {}
for index, row in df1.iterrows():

    func = df1.loc[index]*df1.loc[index].index.astype('float')
    x    = df1.loc[index].index.astype('float')

    dict_INT[index] = integrate.trapz(func, x)

df_out = pd.DataFrame(dict_INT, index=['INT']).T

df_fin = df_out/(a*b)

这是我每行获得的最终金额：

1  3.505796e+06
2  3.068796e+06
3  2.700446e+06
4  2.199336e+06
5  1.840992e+06

Answer 1

我首先通过将数据帧转换为dict，然后按行中的每个项目执行方程式，然后使用集合defaultdict将这些值写入dict，从而解决了这一问题。我将其分解：

import pandas as pd
from collections import defaultdict

df1 = pd.DataFrame({'350': [7.898167, 6.912074, 6.049002, 5.000357, 4.072320],
                '351': [8.094912, 7.090584, 6.221289, 5.154516, 4.211746],
                '352': [8.291657, 7.269095, 6.393576, 5.308674, 4.351173],
                '353': [8.421007, 7.374317, 6.496641, 5.403691, 4.439815],
                '354': [8.535562, 7.463452, 6.584512, 5.485725, 4.517310],
                '355': [8.650118, 7.552586, 6.672383, 4.517310, 4.594806]},
                index=[1, 2, 3, 4, 5]
                 )

int_range = df1.columns.astype(float)
a = 0.005
b = 0.837
dx = 1
df_dict = df1.to_dict() # convert df to dict for easier operations

integrated_dict = {} # initialize empty dict

d = defaultdict(list) # initialize empty dict of lists for tuples later
integrated_list = []
for k,v in df_dict.items(): # unpack df dict of dicts
    for x,y in v.items(): # unpack dicts by column and index (x is index, y is column)
        integrated_list.append((k, (((float(k)*float(y)*float(dx))/(a*b))))) #store a list of tuples.


for x,y in integrated_list: # create dict with column header as key and new integrated calc as value (currently a tuple)
    d[x].append(y)


d = {k:tuple(v) for k, v in d.items()} # unpack to multiple values

integrated_df = pd.DataFrame.from_dict(d) # to df
integrated_df['Sum'] = integrated_df.iloc[:, :].sum(axis=1)

输出（已更新为包括和）：

             350            351            352            353            354  \
0  660539.653524  678928.103226  697410.576822  710302.382557  722004.527599   
1  578070.704898  594694.141935  611402.972521  622015.269056  631317.086738   
2  505890.250896  521785.529032  537763.142652  547984.294624  556969.473835   
3  418189.952210  432314.245161  446512.126165  455795.202628  464025.483871   
4  340576.344086  353243.212903  365976.797133  374493.356033  382109.376344   

         355             Sum
0  733761.502987  4.202947e+06
1  640661.416965  3.678162e+06
2  565996.646356  3.236389e+06
3  383188.781362  2.600026e+06
4  389762.516129  2.206162e+06

熊猫在每一行的列数上进行积分

1 个答案: