如何從 pandas dataframe 轉換為字典

Question

我查看了將 Pandas DataFrame 轉換為字典的指南，以獲取將我的 dataframe 轉換為字典的指南。 但是，我似乎無法更改代碼以將 output 轉換為字典。

以下是我的代碼。

import pandas as pd
import collections


governmentcsv = pd.read_csv('government-procurement-via-gebiz.csv',parse_dates=True) #read csv and it contain dates (parse_dates = true)
extract = governmentcsv.loc[:, ['supplier_name','award_date']] #only getting these columns

extract.award_date= pd.to_datetime(extract.award_date)

def extract_supplier_not_na_2015():
    notNAFifteen = extract[(extract.supplier_name != 'na') & (extract.award_date.dt.year == 2015)] #extract only year 2015
    notNAFifteen.reset_index(drop = True,inplace=True) #reset index
    notNAFifteen.index += 1 #and index start from 1
    #SortednotNAFifteen = collections.orderedDictionary(notNAFifteen)

    return notNAFifteen

print extract_supplier_not_na_2015()

OUTPUT 是：

                                          supplier_name award_date
1                               RMA CONTRACTS PTE. LTD. 2015-01-28
2     TESCOM (SINGAPORE) SOFTWARE SYSTEMS TESTING PT... 2015-07-01
3                                  MKS GLOBAL PTE. LTD. 2015-04-24
4               CERTIS TECHNOLOGY (SINGAPORE) PTE. LTD. 2015-06-26
5                    RHT COMPLIANCE SOLUTIONS PTE. LTD. 2015-08-14
6                                   CLEANMAGE PTE. LTD. 2015-07-30
7                             SOLUTIONSATWORK PTE. LTD. 2015-11-23
8                                       Ufinity Pte Ltd 2015-05-04
9                                         NCS PTE. LTD. 2015-01-28

Answer 1

我想我找到了這個數據集： https://data.gov.sg/dataset/government-procurement

無論如何，這是代碼

import pandas as pd


df = pd.read_csv('government-procurement-via-gebiz.csv', 
                  encoding='unicode_escape', 
                  parse_dates=['award_date'], 
                  infer_datetime_format=True,
                  usecols=['supplier_name', 'award_date'], 
)

df = df[(df['supplier_name'] != 'Unknown') & (df['award_date'].dt.year == 2015)].reset_index(drop=True)

#Faster way:
d1 = df.set_index('supplier_name').to_dict()['award_date']

#Alernatively:
d2 = dict(zip(df['supplier_name'], df['award_date']))