Evaluating multiple conditions in if-then-else block in a Pandas DataFrame

Question

I want to create a new column in a Pandas DataFrame by evaluating multiple conditions in an if-then-else block.

if events.hour <= 6:
    events['time_slice'] = 'night'
elif events.hour <= 12:
    events['time_slice'] = 'morning'
elif events.hour <= 18:
    events['time_slice'] = 'afternoon'
elif events.hour <= 23:
    events['time_slice'] = 'evening'

When I run this, I get the error below:

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

So I tried to solve this by adding the any statement like shown below:

if (events.hour <= 6).any():
    events['time_slice'] = 'night'
elif (events.hour <= 12).any():
    events['time_slice'] = 'morning'
elif (events.hour <= 18).any():
    events['time_slice'] = 'afternoon'
elif (events.hour <= 23).any():
    events['time_slice'] = 'evening'

Now I do not get any error, but when I check the unique values of time_slice, it only shows 'night'

np.unique(events.time_slice)

array(['night'], dtype=object)

How can I solve this, because my data contains samples that should get 'morning', 'afternoon' or 'evening'. Thanks!

MaxU - stand with Ukraine · Accepted Answer · 2016-07-23 09:46:20Z

4

you can use pd.cut() method in order to categorize your data:

Demo:

In [66]: events = pd.DataFrame(np.random.randint(0, 23, 10), columns=['hour'])

In [67]: events
Out[67]:
   hour
0     5
1    17
2    12
3     2
4    20
5    22
6    20
7    11
8    14
9     8

In [71]: events['time_slice'] = pd.cut(events.hour, bins=[-1, 6, 12, 18, 23], labels=['night','morning','afternoon','evening'])

In [72]: events
Out[72]:
   hour time_slice
0     5      night
1    17  afternoon
2    12    morning
3     2      night
4    20    evening
5    22    evening
6    20    evening
7    11    morning
8    14  afternoon
9     8    morning

edited Jul 23, 2016 at 9:46

answered Jul 23, 2016 at 8:52

MaxU - stand with Ukraine

212k37 gold badges402 silver badges437 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Divakar Over a year ago

Maybe the first bin would be -1 because if the entry is 0, it might make it NaN?

Julien · Accepted Answer · 2016-07-23 08:55:23Z

2

You could create a function:

def time_slice(hour):
    if hour <= 6:
        return 'night'
    elif hour <= 12:
        return 'morning'
    elif hour <= 18:
        return 'afternoon'
    elif hour <= 23:
        return 'evening'

then events['time_slice'] = events.hour.apply(time_slice) should do the trick.

answered Jul 23, 2016 at 8:55

Julien

15.3k6 gold badges33 silver badges58 bronze badges

Comments

Divakar · Accepted Answer · 2016-07-23 09:27:26Z

2

Here's a NumPy approach to it -

tags = ['night','morning','afternoon','evening']
events['time_slice'] = np.take(tags,((events.hour.values-1)//6).clip(min=0))

Sample run -

In [130]: events
Out[130]: 
   hour time_slice
0     0      night
1     8    morning
2    16  afternoon
3    20    evening
4     2      night
5    14  afternoon
6     7    morning
7    18  afternoon
8     8    morning
9    22    evening

edited Jul 23, 2016 at 9:27

answered Jul 23, 2016 at 9:08

Divakar

222k19 gold badges273 silver badges374 bronze badges

2 Comments

MaxU - stand with Ukraine Over a year ago

((events.hour.values-1)//6).clip(min=0) - this is pretty smart

Divakar Over a year ago

@MaxU I guess so, but works for regular intervals only.

Collectives™ on Stack Overflow

Evaluating multiple conditions in if-then-else block in a Pandas DataFrame

3 Answers 3

1 Comment

Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related