Reputation: 151
I have a data frame like this:
BuyTime ID SellTime
94650 1 94651
94717 1 94817
120458 2 114119
the buy time and sell time type are integer but i want to convert them to standard time date. I already used
quote['SellTime'] = pd.to_datetime(quote['SellTime'], unit = 's')
but it gives a start year like 1970-01-01 to time which is not needed. my data should be like this:
BuyTime Id SellTime
09:46:50 1 09:46:51
09:47:17 1 09:48:17
12:04:58 2 11:41:19
EDIT: also using this function regard to size of my data is not efficient:
def format_time(value):
t2 = str(value)
if len(t2) == 5:
return "0%s:%s:%s" % (t2[0], t2[1:3], t2[3:5])
return "%s:%s:%s" % (t2[0:2], t2[2:4], t2[4:6])
Upvotes: 3
Views: 2699
Reputation: 862801
I think you need if need output as string
s zfill
and str[]
for select by positions:
t1 = quote['BuyTime'].astype(str).str.zfill(6)
t2 = quote['SellTime'].astype(str).str.zfill(6)
quote['BuyTime'] = t1.str[0:2] + ':' + t1.str[2:4] + ':' + t1.str[4:6]
quote['SellTime'] = t2.str[0:2] + ':' + t2.str[2:4] + ':' + t2.str[4:6]
print (quote)
BuyTime ID SellTime
0 09:46:50 1 09:46:51
1 09:47:17 1 09:48:17
2 12:04:58 2 11:41:19
Or if need python times
add 0
by zfill
, convert to datetime
s and extract time
s:
t1 = quote['BuyTime'].astype(str).str.zfill(6)
t2 = quote['SellTime'].astype(str).str.zfill(6)
quote['BuyTime'] = pd.to_datetime(t1, format='%H%M%S').dt.time
quote['SellTime'] = pd.to_datetime(t2, format='%H%M%S').dt.time
print (quote)
BuyTime ID SellTime
0 09:46:50 1 09:46:51
1 09:47:17 1 09:48:17
2 12:04:58 2 11:41:19
Alternative for string
s outputs is strftime
:
quote['BuyTime'] = pd.to_datetime(t1, format='%H%M%S').dt.strftime('%H:%M:%S')
quote['SellTime'] = pd.to_datetime(t2, format='%H%M%S').dt.strftime('%H:%M:%S')
print (quote)
BuyTime ID SellTime
0 09:46:50 1 09:46:51
1 09:47:17 1 09:48:17
2 12:04:58 2 11:41:19
Upvotes: 3
Reputation: 1922
Assuming the date string only contains hour/minute/second (and minute/second are always 2 characters long, you could do something like this to parse the string using Python's modulo operator:
dateStr = '94522'
dateInt = int(dateStr)
sec = dateInt % 100
dateInt = dateInt - sec
min = dateInt % 10000
dateInt = dateInt - min
min /= 100
hour = dateInt / 10000
newDateStr = '{}:{}:{}'.format(hour, min, sec)
print 'number={} formatted={}'.format(dateStr, newDateStr)
This will work for both 5 and 6 character long time strings.
Upvotes: 0