How to convert string into datetime?

Question

I'm quite new to Python and I'm encountering a problem.

I have a dataframe where one of the columns is the departure time of flights. These hours are given in the following format : 1100.0, 525.0, 1640.0, etc.

This is a pandas series which I want to transform into a datetime series such as : S = [11.00, 5.25, 16.40,...]

What I have tried already :

Transforming my objects into string :

S = [str(x) for x in S]

Using datetime.strptime :

S =  [datetime.strptime(x,'%H%M.%S') for x in S]

But since they are not all the same format it doesn't work

Using parser from dateutil :

S =  [parser.parse(x) for x in S]

I got the error :

 'Unknown string format'

Using the panda datetime :

S= pd.to_datetime(S)

Doesn't give me the expected result

Thanks for your answers !

Why don't you split on the point and then check the string length? If it is 3, then do datetime (0,0,0,int(srting_before_point[0]),int(string_before_point[-2:]),after_point_string) and something similar if the length is 4. — Nerdrigo
– Nerdrigo, Commented Apr 3, 2019 at 16:47
can you show us an example that datetime.strptime(x,'%H%M.%S') does not work ? — Mojtaba Kamyabi
– Mojtaba Kamyabi, Commented Apr 3, 2019 at 16:49
@Nerdrigo Wouldn't it be far easier to just divide each element by 100? — Luke
– Luke, Commented Apr 3, 2019 at 16:50
So these are initially floating point numbers? Why not just do the simple math to separate out hours and minutes, and then pass those numbers to a datetime constructor. I don't see any reason to do string operations here. — CryptoFool
– CryptoFool, Commented Apr 3, 2019 at 16:50
Not a solution but a way to normalize the formats: s = [f'{x:07.2f}' for x in s]. That will make s = ['1100.00', '0525.00', '1640.00'] for s = [1100.0, 525.0, 1640.0]. — accdias
– accdias, Commented Apr 3, 2019 at 16:57

Rocky Li · Accepted Answer · 2019-04-03 16:52:44Z

1

Since it's a columns within a dataframe (A series), keep it that way while transforming should work just fine.

S = [1100.0, 525.0, 1640.0]
se = pd.Series(S) # Your column

# se:
0    1100.0
1     525.0
2    1640.0
dtype: float64

setime = se.astype(int).astype(str).apply(lambda x: x[:-2] + ":" + x[-2:])

This transform the floats to correctly formatted strings:

0    11:00
1     5:25
2    16:40
dtype: object

And then you can simply do:

df["your_new_col"] = pd.to_datetime(setime)

answered Apr 3, 2019 at 16:52

Rocky Li

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Jimmy · Accepted Answer · 2019-04-03 17:14:34Z

0

How about this?

(Added an if statement since some entries have 4 digits before decimal and some have 3. Added the use case of 125.0 to account for this)

from datetime import datetime

S = [1100.0, 525.0, 1640.0, 125.0]

for x in S: if str(x).find(".")==3: x="0"+str(x) print(datetime.strftime(datetime.strptime(str(x),"%H%M.%S"),"%H:%M:%S"))

edited Apr 3, 2019 at 17:14

answered Apr 3, 2019 at 17:00

Jimmy

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Comments

Konstantin Grigorov · Accepted Answer · 2019-04-03 17:16:33Z

You might give it a go as follows:

# Just initialising a state in line with your requirements
st = ["1100.0", "525.0", "1640.0"]
dfObj = pd.DataFrame(st)

# Casting the string column to float
dfObj_num = dfObj[0].astype(float)

# Getting the hour representation out of the number 
df1 = dfObj_num.floordiv(100)

# Getting the minutes
df2 = dfObj_num.mod(100)

# Moving the minutes on the right-hand side of the decimal point
df3 = df2.mul(0.01)

# Combining the two dataframes
df4 = df1.add(df3)

# At this point can cast to other types

Result:

0    11.00
1     5.25
2    16.40

You can run this example to verify the steps for yourself, also you can make it into a function. Make slight variations if needed in order to tweak it according to your precise requirements.

Might be useful to go through this article about Pandas Series. https://www.geeksforgeeks.org/python-pandas-series/

bart cubrich · Accepted Answer · 2019-04-03 17:29:27Z

0

There must be a better way to do this, but this works for me.

df=pd.DataFrame([1100.0, 525.0, 1640.0], columns=['hour'])
df['hour_dt']=((df['hour']/100).apply(str).str.split('.').str[0]+'.'+
  df['hour'].apply((lambda x: '{:.2f}'.format(x/100).split('.')[1])).apply(str))

print(df)

     hour hour_dt
0  1100.0   11.00
1   525.0    5.25
2  1640.0   16.40

edited Apr 3, 2019 at 17:29

answered Apr 3, 2019 at 17:16

bart cubrich

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