如何從嵌套字典中提取值作為 Pandas DataFrames

Question

你好stackoverflowers，

我希望您能幫助我理解嵌套字典遇到的問題。 我從 excel 文件中抓取了一些表格： ['Table 5','Table 8',Table 40'] 。 我從我使用的代碼中得到的是一個嵌套字典，我不確定如何處理。 我想這些才是初學者真正的痛苦。 我的目標是使用鍵將值轉換為數據框（例如表 5）。 原表：

數據框示例：

d = {0: ['TB','VT','BT','CI','CH','CL','RT','RU','PV','PV','PV','PV','PV','RH','PV','PV','PV','PV','NaN','NaN','TB','VT','BT','CI','CH','CL','RT','RU','PV','PV'], 
     1: ['Table 1','BRAND. SUMMARY','Base: Floating Base (TOTAL) (18-59)','NaN','NaN','NaN','Base','Unweighted row','brand1','brand2','brand3','brand4','NPS','','NaN','Row1','Row2','Row3','NaN','NaN','Table 5','Brands Title 1','Base: All (TOTAL) (18-59)','NaN','NaN','NaN','Base','Unweighted row','Brand1','Brand2'],
     2: ['NaN','NaN','NaN','(TOTAL)','Discrete monthly banner','Sept (a)','100','997','0.31','0.31','0.31','0.31','0.31','NaN','0.62','0.64','0.61','0.6','NaN','NaN','NaN','NaN','NaN','NaN','NaN','Total','19479','19608','0.75','0.75'],
     3: ['NaN','NaN','NaN','NaN','NaN','Oct (b)','1090','1100','0.31','0.31','0.31','0.31','0.31','NaN','0.64','0.67','0.64','0.64','NaN','NaN','NaN','NaN','NaN','TOTAL','Discrete monthly banner','Sept (a)','1000','1000','0.8','0.8'],
     4: ['NaN','NaN','NaN','NaN','NaN','Nov (c)','3164','3191','0.31','0.31','0.31','0.31','0.31','NaN','0.64','0.67','0.64','0.64','NaN','NaN','NaN','NaN','NaN','NaN','NaN','Oct (b)','1000','1000','0.8','0.8'],
     5: ['NaN','NaN','NaN','NaN','NaN','Dec (d)','992','3999','0.31','0.31','0.31','0.31','0.31','NaN','0.51','0.47','0.67','0.61','NaN','NaN','NaN','NaN','NaN','NaN','NaN','Nov (c)','1000','1000','0.8','0.8']}

當我打印表值和鍵時，會返回：

第 174 行應該是我的列標題。

這是我用來從 Excel 中抓取表格的代碼：

ws = pd.read_excel(r'C:\Users\Tables.xlsx', sheet_name= "Percents", header = None, usecols="B:XFD")

table_names = ["Table 5", "Table 8", "Table 9", "Table 40"]
groups = ws[1].isin(table_names).cumsum()
tables = {g.iloc[0,0]: g.iloc[1:24] for k,g in ws.groupby(groups)}
#because the syntax above (e.g.tables={g.iloc}) returned also the other values, I filtered again based on the table names
filtered_d = dict((k, tables[k]) for k in table_names if k in tables)

我嘗試修改此代碼以返回我的值，但是當我刪除orient="index"或說orient="columns"我收到錯誤消息。 我認為 for 循環可以解決問題。

df = pd.DataFrame.from_dict({(i,j): filtered_d[i][j] 
                           for i in filtered_d.keys() 
                           for j in filtered_d[i].keys()}, orient="index")

如何通過保持當前表格格式並將每個值轉換為數據框來解決這個問題？

預先感謝您給我的任何建議。

Answer 1

我不完全確定您想要什么輸出，但是通過提供的示例，我們可以試一試。 這是你追求的嗎？

import pandas as pd
df = pd.DataFrame({0: ['TB','VT','BT','CI','CH','CL','RT','RU','PV','PV','PV','PV','PV','RH','PV','PV','PV','PV','NaN','NaN','TB','VT','BT','CI','CH','CL','RT','RU','PV','PV'], 
     1: ['Table 1','BRAND. SUMMARY','Base: Floating Base (TOTAL) (18-59)','NaN','NaN','NaN','Base','Unweighted row','brand1','brand2','brand3','brand4','NPS','','NaN','Row1','Row2','Row3','NaN','NaN','Table 5','Brands Title 1','Base: All (TOTAL) (18-59)','NaN','NaN','NaN','Base','Unweighted row','Brand1','Brand2'],
     2: ['NaN','NaN','NaN','(TOTAL)','Discrete monthly banner','Sept (a)','100','997','0.31','0.31','0.31','0.31','0.31','NaN','0.62','0.64','0.61','0.6','NaN','NaN','NaN','NaN','NaN','NaN','NaN','Total','19479','19608','0.75','0.75'],
     3: ['NaN','NaN','NaN','NaN','NaN','Oct (b)','1090','1100','0.31','0.31','0.31','0.31','0.31','NaN','0.64','0.67','0.64','0.64','NaN','NaN','NaN','NaN','NaN','TOTAL','Discrete monthly banner','Sept (a)','1000','1000','0.8','0.8'],
     4: ['NaN','NaN','NaN','NaN','NaN','Nov (c)','3164','3191','0.31','0.31','0.31','0.31','0.31','NaN','0.64','0.67','0.64','0.64','NaN','NaN','NaN','NaN','NaN','NaN','NaN','Oct (b)','1000','1000','0.8','0.8'],
     5: ['NaN','NaN','NaN','NaN','NaN','Dec (d)','992','3999','0.31','0.31','0.31','0.31','0.31','NaN','0.51','0.47','0.67','0.61','NaN','NaN','NaN','NaN','NaN','NaN','NaN','Nov (c)','1000','1000','0.8','0.8']})
tbl = df.drop(range(5), axis=0).drop(0, axis=1)
print(tbl)

或者，您可能想適當地命名行和列：

index = tbl.iloc[:,0]
columns = tbl.iloc[0]
data = df.drop(range(6), axis=0).drop(range(2), axis=1)
tbl2 = pd.DataFrame(data, index=index, columns=columns)

無論如何，希望您可以將其強制為正確的格式。