'TypeError: strptime() argument 1 must be str, not float
I'm having parsing errors on my code, below is the code and almost understandable dataset
import numpy as np
import pandas as pd
from datetime import datetime as dt
data0 = pd.read_csv('2009-10.csv')
data1 = pd.read_csv('2010-11.csv')
def parse_date(date):
if date == '':
return None
else:
return dt.strptime(date, '%d/%m/%y').date()
data0.Date = data0.Date.apply(parse_date)
data1.Date = data1.Date.apply(parse_date)
TypeError: strptime() argument 1 must be str, not float
Date HomeTeam AwayTeam FTHG FTAG FTR HTHG HTAG HTR Referee HS AS HST AST HF AF HC AC HY AY HR AR B365H B365D B365A
15/08/09 Aston Villa Wigan 0 2 A 0 1 A M Clattenburg 11 14 5 7 15 14 4 6 2 2 0 0 1.67 3.6 5.5
15/08/09 Blackburn Man City 0 2 A 0 1 A M Dean 17 8 9 5 12 9 5 4 2 1 0 0 3.6 3.25 2.1
15/08/09 Bolton Sunderland 0 1 A 0 1 A A Marriner 11 20 3 13 16 10 4 7 2 1 0 0 2.25 3.25 3.25
15/08/09 Chelsea Hull 2 1 H 1 1 D A Wiley 26 7 12 3 13 15 12 4 1 2 0 0 1.17 6.5 21
15/08/09 Everton Arsenal 1 6 A 0 3 A M Halsey 8 15 5 9 11 13 4 9 0 0 0 0 3.2 3.25 2.3
Date HomeTeam AwayTeam FTHG FTAG FTR HTHG HTAG HTR Referee HS AS HST AST HF AF HC AC HY AY HR AR B365H B365D B365A
14/08/10 Aston Villa West Ham 3 0 H 2 0 H M Dean 23 12 11 2 15 15 16 7 1 2 0 0 2 3.3 4
14/08/10 Blackburn Everton 1 0 H 1 0 H P Dowd 7 17 2 12 19 14 1 3 2 1 0 0 2.88 3.25 2.5
14/08/10 Bolton Fulham 0 0 D 0 0 D S Attwell 13 12 9 7 12 13 4 8 1 3 0 0 2.2 3.3 3.4
14/08/10 Chelsea West Brom 6 0 H 2 0 H M Clattenburg 18 10 13 4 10 10 3 1 1 0 0 0 1.17 7 17
14/08/10 Sunderland Birmingham 2 2 D 1 0 H A Taylor 6 13 2 7 13 10 3 6 3 3 1 0 2.1 3.3 3.6
14/08/10 Tottenham Man City 0 0 D 0 0 D A Marriner 22 11 18 7 13 16 10 3 0 2 0 0 2.4 3.3 3
Solution 1:[1]
IIUC, I think you are converting strings into datetime dtypes.
You can use Pandas to_datetime
:
data0['Date'] = pd.to_datetime(data0['Date'], format='%d/%m/%y')
data1['Date'] = pd.to_datetime(data1['Date'], format='%d/%m/%y')
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | Scott Boston |