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
我想解决一个方程，如下图所示：
I等于数据帧中的值。 x是int_range值，因此在这种情况下，dx = 1时从350到355。
a和b是可选常数
我需要获取一个数据框作为每行的输出
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)
这是我每行得到的最终总和：
1 3.505796e+06
2 3.068796e+06
3 2.700446e+06
4 2.199336e+06
5 1.840992e+06
最佳答案
通过首先将数据帧转换为dict，然后按行中的每个项目执行方程式，然后使用collectiondefaultdict将这些值写入dict，来解决此问题。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（更新为包括总和）：
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
关于python - Pandas 在每一行的列数上进行积分，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/52370693/