unsupported operand type(s) for +: 'float'

Question

I have a dataset with one input with date and time . First I wrote the code to find the first time of 5 value in X3 column and I turn that time into 0. Then I tried to add timedelta(hours=1) into that which is having range 6. Then it gave me this error "unsupported operand type(s) for +: 'bool' and 'datetime.timedelta'"

Can anyone help me to solve this error?

my code:

data =pd.read_csv('data6.csv')

data['time_diff']= pd.to_datetime(data['date'] + " " + data['time'],
               format='%d/%m/%Y %H:%M:%S', dayfirst=True)
mask = data['X3'].eq(5)
data['duration'] = data[mask].drop_duplicates(['date','X3']).groupby(['date','X3'])['time_diff'].transform('first')
data['duration'] = data['time_diff'].sub(data['duration']).dt.total_seconds().div(3600)

date    time    x3            Time(expected)
10/3/2018   6:15:00     0        NaN
10/3/2018   6:45:00     5        0.0
10/3/2018   7:45:00     0        NaN
10/3/2018   9:00:00     0        NaN
10/3/2018   9:25:00     7        NaN
10/3/2018   9:30:00     0        NaN
10/3/2018   11:00:00    0        NaN
10/3/2018   11:30:00    0        NaN
10/3/2018   13:30:00    0        NaN
10/3/2018   13:50:00    5        NaN
10/3/2018   15:00:00    0        NaN
10/3/2018   15:25:00    0        NaN
10/3/2018   16:25:00    0        NaN
10/3/2018   18:00:00    7        NaN
10/3/2018   19:00:00    0        NaN
10/3/2018   19:30:00    0        NaN
10/3/2018   20:00:00    0        NaN
10/3/2018   22:05:00    0        NaN
10/3/2018   22:15:00    5        NaN
10/3/2018   23:40:00    0        NaN
10/4/2018   6:58:00     5        0.0
10/4/2018   13:00:00    0        NaN
10/4/2018   16:00:00    7        NaN
10/4/2018   17:00:00    7        NaN

So here I have a summation equation to apply for X3 column value.

Then according to this summation equation I want to take the value of X3 in every hour.

That's why first I found the start time of mention value of 5 in every day and then convert that time into 0:00:00.

Then from that start time adding one hour one hour till to 6 hour I need to take the value for A.

For A equation is :

A =  X3(5) - M

So first I took the first time separately, 0 time only. For that I used the code :

time= data['duration'].eq(0)

Then I wrote this equation method inside the class

time=0
M=0
for _ in range(len(data['X3'])):
  if X3.all()==5:
    if time ==data['duration'].eq(5).all():
        M=X3
        for i in (time + timedelta(hours=1*it) for it in range(6)):
            M = 5 - 0.0015 * np.sum(i*X3) 
print(M)   

Then got the values only 0 .

Then this error came .

From these code what I am expecting output is:

					
	   time	 	                      expected output		
	0 (start time of x3 value of 5)         5		
	1 hr		                   5-0.3(according to the summation equation) = 4.7		
	2hr		                      5-0.6=4.4		
	3hr		                      5-0.9=4.1		
	4hr		                      5-1.2=3.8		
	5hr		                      5-1.5=3.5		
	6hr		                      5-1.8=3.2		

Answer 1

In your code time= data['duration'].eq(0) this part is making the time variable of bool type. Try to convert it to 0 or 1 then add it in for loop.

If time is scalar (single) value :

if (time==True):
    time=1
else:
    time=0

if time is vector (array) :

time_array = [0 if tm==False else 1 for tm in data['duration'].eq(0)]

Also, there is one more mistake in your for loop. You can't add 1 like this to data time. Try this :

from datetime import datetime, timedelta
current_time = datetime.now()
nine_hours_from_now = current_time  + timedelta(hours=9)

In this code you can replace timedelta(hours=9) by your time timedelta(hours = time).

Note : your time will be a binary array. you have to reaverse is to pick the values and add.