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Pandas change time values based on condition

I have a dataframe:
data = {'time':['08:45:00', '09:30:00', '18:00:00', '15:00:00']}
df = pd.DataFrame(data)
I would like to convert the time based on conditions: if the hour is less than 9, I want to set it to 9 and if the hour is more than 17, I need to set it to 17.
I tried this approach:
df['time'] = np.where(((df['time'].dt.hour < 9) & (df['time'].dt.hour != 0)), dt.time(9, 00))
I am getting an error: Can only use .dt. accesor with datetimelike values.
Can anyone please help me with this? Thanks.

Here's a way to do what your question asks:
df.time = pd.to_datetime(df.time)
df.loc[df.time.dt.hour < 9, 'time'] = (df.time.astype('int64') + (9 - df.time.dt.hour)*3600*1000000000).astype('datetime64[ns]')
df.loc[df.time.dt.hour > 17, 'time'] = (df.time.astype('int64') + (17 - df.time.dt.hour)*3600*1000000000).astype('datetime64[ns]')
Input:
time
0 2022-06-06 08:45:00
1 2022-06-06 09:30:00
2 2022-06-06 18:00:00
3 2022-06-06 15:00:00
Output:
time
0 2022-06-06 09:45:00
1 2022-06-06 09:30:00
2 2022-06-06 17:00:00
3 2022-06-06 15:00:00
UPDATE:
Here's alternative code to try to address OP's error as described in the comments:
import pandas as pd
import datetime
data = {'time':['08:45:00', '09:30:00', '18:00:00', '15:00:00']}
df = pd.DataFrame(data)
print('', 'df loaded as strings:', df, sep='\n')
df.time = pd.to_datetime(df.time, format='%H:%M:%S')
print('', 'df converted to datetime by pd.to_datetime():', df, sep='\n')
df.loc[df.time.dt.hour < 9, 'time'] = (df.time.astype('int64') + (9 - df.time.dt.hour)*3600*1000000000).astype('datetime64[ns]')
df.loc[df.time.dt.hour > 17, 'time'] = (df.time.astype('int64') + (17 - df.time.dt.hour)*3600*1000000000).astype('datetime64[ns]')
df.time = [time.time() for time in pd.to_datetime(df.time)]
print('', 'df with time column adjusted to have hour between 9 and 17, converted to type "time":', df, sep='\n')
Output:
df loaded as strings:
time
0 08:45:00
1 09:30:00
2 18:00:00
3 15:00:00
df converted to datetime by pd.to_datetime():
time
0 1900-01-01 08:45:00
1 1900-01-01 09:30:00
2 1900-01-01 18:00:00
3 1900-01-01 15:00:00
df with time column adjusted to have hour between 9 and 17, converted to type "time":
time
0 09:45:00
1 09:30:00
2 17:00:00
3 15:00:00
UPDATE #2:
To not just change the hour for out-of-window times, but to simply apply 9:00 and 17:00 as min and max times, respectively (see OP's comment on this), you can do this:
df.loc[df['time'].dt.hour < 9, 'time'] = pd.to_datetime(pd.DataFrame({
'year':df['time'].dt.year, 'month':df['time'].dt.month, 'day':df['time'].dt.day,
'hour':[9]*len(df.index)}))
df.loc[df['time'].dt.hour > 17, 'time'] = pd.to_datetime(pd.DataFrame({
'year':df['time'].dt.year, 'month':df['time'].dt.month, 'day':df['time'].dt.day,
'hour':[17]*len(df.index)}))
df['time'] = [time.time() for time in pd.to_datetime(df['time'])]

Since your 'time' column contains strings they can kept as strings and assign new string values where appropriate. To filter for your criteria it is convenient to: create datetime Series from the 'time' column, create boolean Series by comparing the datetime Series with your criteria, use the boolean Series to filter the rows which need to be changed.
Your data:
import numpy as np
import pandas as pd
data = {'time':['08:45:00', '09:30:00', '18:00:00', '15:00:00']}
df = pd.DataFrame(data)
print(df.to_string())
>>>
time
0 08:45:00
1 09:30:00
2 18:00:00
3 15:00:00
Convert to datetime, make boolean Series with your criteria
dts = pd.to_datetime(df['time'])
lt_nine = dts.dt.hour < 9
gt_seventeen = (dts.dt.hour >= 17)
print(lt_nine)
print(gt_seventeen)
>>>
0 True
1 False
2 False
3 False
Name: time, dtype: bool
0 False
1 False
2 True
3 False
Name: time, dtype: bool
Use the boolean series to assign a new value:
df.loc[lt_nine,'time'] = '09:00:00'
df.loc[gt_seventeen,'time'] = '17:00:00'
print(df.to_string())
>>>
time
0 09:00:00
1 09:30:00
2 17:00:00
3 15:00:00
Or just stick with strings altogether and create the boolean Series using regex patterns and .str.match.
data = {'time':['08:45:00', '09:30:00', '18:00:00', '15:00:00','07:22:00','22:02:06']}
dg = pd.DataFrame(data)
print(dg.to_string())
>>>
time
0 08:45:00
1 09:30:00
2 18:00:00
3 15:00:00
4 07:22:00
5 22:02:06
# regex patterns
pattern_lt_nine = '^00|01|02|03|04|05|06|07|08'
pattern_gt_seventeen = '^17|18|19|20|21|22|23'
Make boolean Series and assign new values
gt_seventeen = dg['time'].str.match(pattern_gt_seventeen)
lt_nine = dg['time'].str.match(pattern_lt_nine)
dg.loc[lt_nine,'time'] = '09:00:00'
dg.loc[gt_seventeen,'time'] = '17:00:00'
print(dg.to_string())
>>>
time
0 09:00:00
1 09:30:00
2 17:00:00
3 15:00:00
4 09:00:00
5 17:00:00
Time series / date functionality
Working with text data

Finding the beginning and end dates of when a sequence of values occurs in Pandas

I have a dataframe with an index column and another column that marks whether or not an event occurred on that day with a 1 or 0.
If an event occurred it typically happened continuously for a prolonged period of time. They'll typically mark whether or not a recession occurred, so it'd likely be 60-180 straight days that would be marked with a 1 before going to 0 again.
What I need to do is find the dates that mark the beginning and end of each sequence of 1's.
Here's some quick sample code:
dates = pd.date_range(start='2010-01-01', end='2015-01-01')
nums = np.random.normal(50, 5, 1827)
df = pd.DataFrame(nums, index=dates, columns=['Nums'])
df['Recession'] = np.where((df.index.month == 3) | (df.index.month == 12), 1, 0)
With the example dataframe, the value 1 occurs for the months of March and December, so ideally I'd have a list that reads [2010-03-01, 2010-03-31, 2010-12-01, 2010-12-30, ......, 2015-12-01, 2015-12-30].
I know I could find these values by using a for-loop, but that seems inefficient. I tried using groupby as well, but couldn't find anything that gave the results that I wanted.
Not sure if there's a pandas or numpy method to search an index for the appropriate conditions or not.

Let's try this, using DataFrameGroupBy.idxmin + DataFrameGroupBy.idxmax
# group-by on month, year & aggregate on date
g = (
df.assign(day=df.index.day)
.groupby([df.index.month, df.index.year]).day
)
# create mask of max date & min date for each (month, year) combination
mask = df.index.isin(g.idxmin()) | df.index.isin(g.idxmax())
# apply previous mask with month filter..
df.loc[mask & (df.index.month.isin([3,12])), 'Recession'] = 1
print(df[df['Recession'] == 1])
Nums Recession
2010-03-01 45.698168 1.0
2010-03-31 47.969167 1.0
2010-12-01 49.388595 1.0
2010-12-31 46.689064 1.0
2011-03-01 50.120603 1.0
2011-03-31 58.379980 1.0
2011-12-01 53.745407 1.0
...
...

I would use diff to find the periods, the diff enables to find when it switches from one state to another, so split the indices found in two parts, the starts and ends.
Depending whether the data starts with a recession or not:
locs = (df.Recession.diff().fillna(0)!=0).values.nonzero()[0]
if df.Recession.iloc[0]==0:
start = df.index[locs[::2]]
end = df.index[locs[1::2]-1]
else:
start = df.index[locs[::2]-1]
end = df.index[locs[1::2]]
If the data started with a recession already, up to you if you want to include the first date as a start or not, the above does not include it.

From what I understand you need to find the first value in a sequence? if so we can use groupby and cumsum to sum each consecutive group, and cumcount to count each of the groups.
df["keyGroup"] = (
df.groupby(df["Recession"].ne(df["Recession"].shift()).cumsum()).cumcount() + 1
)
df[df['keyGroup'].eq(1)]
Nums Recession keyGroup
2010-01-01 51.944742 0 1
2010-03-01 54.809271 1 1
2010-04-01 52.632831 0 1
2010-12-01 55.863695 1 1
2011-01-01 52.944778 0 1
2011-03-01 58.164943 1 1
2011-04-01 49.590640 0 1
2011-12-01 47.884919 1 1
2012-01-01 44.128065 0 1
2012-03-01 54.846231 1 1
2012-04-01 51.312064 0 1
2012-12-01 46.091171 1 1
2013-01-01 49.287102 0 1
2013-03-01 54.727874 1 1
2013-04-01 53.163730 0 1
2013-12-01 42.373602 1 1
2014-01-01 43.822791 0 1
2014-03-01 51.203125 1 1
2014-04-01 54.322415 0 1
2014-12-01 44.052536 1 1
2015-01-01 53.438015 0 1
you can call .index to get the values in a list.
df[df['keyGroup'].eq(1)].index
DatetimeIndex(['2010-01-01', '2010-03-01', '2010-04-01', '2010-12-01',
'2011-01-01', '2011-03-01', '2011-04-01', '2011-12-01',
'2012-01-01', '2012-03-01', '2012-04-01', '2012-12-01',
'2013-01-01', '2013-03-01', '2013-04-01', '2013-12-01',
'2014-01-01', '2014-03-01', '2014-04-01', '2014-12-01',
'2015-01-01'],
dtype='datetime64[ns]', name='date', freq=None)

Comparing date column values in one dateframe with two date column in another dataframe by row in Pandas

I have a dataframe like this with two date columns and a quamtity column :
start_date end_date qty
1 2018-01-01 2018-01-08 23
2 2018-01-08 2018-01-15 21
3 2018-01-15 2018-01-22 5
4 2018-01-22 2018-01-29 12
I have a second dataframe with just column containing yearly holidays for a couple of years, like this:
holiday
1 2018-01-01
2 2018-01-27
3 2018-12-25
4 2018-12-26
I would like to go through the first dataframe row by row and assign boolean value to a new column holidays if a date in the second data frame falls between the date values of the first date frame. The result would look like this:
start_date end_date qty holidays
1 2018-01-01 2018-01-08 23 True
2 2018-01-08 2018-01-15 21 False
3 2018-01-15 2018-01-22 5 False
4 2018-01-22 2018-01-29 12 True
When I try to do that with a for loop I get the following error:
ValueError: Can only compare identically-labeled Series objects
An answer would be appreciated.

If you want a fully-vectorized solution, consider using the underlying numpy arrays:
import numpy as np
def holiday_arr(start, end, holidays):
start = start.reshape((-1, 1))
end = end.reshape((-1, 1))
holidays = holidays.reshape((1, -1))
result = np.any(
(start <= holiday) & (holiday <= end),
axis=1
)
return result
If you have your dataframes as above (calling them df1 and df2), you can obtain your desired result by running:
df1["contains_holiday"] = holiday_arr(
df1["start_date"].to_numpy(),
df1["end_date"].to_numpy(),
df2["holiday"].to_numpy()
)
df1 then looks like:
start_date end_date qty contains_holiday
1 2018-01-01 2018-01-08 23 True
2 2018-01-08 2018-01-15 21 False
3 2018-01-15 2018-01-22 5 False
4 2018-01-22 2018-01-29 12 True

try:
def _is_holiday(row, df2):
return ((df2['holiday'] >= row['start_date']) & (df2['holiday'] <= row['end_date'])).any()
df1.apply(lambda x: _is_holiday(x, df2), axis=1)

I'm not sure why you would want to go row-by-row. But boolean comparisons would be way faster.
df['holiday'] = ((df2.holiday >= df.start_date) & (df2.holiday <= df.end_date))
Time
>>> 1000 loops, best of 3: 1.05 ms per loop
Quoting #hchw solution (row-by-row)
def _is_holiday(row, df2):
return ((df2['holiday'] >= row['start_date']) & (df2['holiday'] <= row['end_date'])).any()
df.apply(lambda x: _is_holiday(x, df2), axis=1)
>>> The slowest run took 4.89 times longer than the fastest. This could mean that an intermediate result is being cached.
100 loops, best of 3: 4.46 ms per loop

Try IntervalIndex.contains with list comprehensiont and np.sum
iix = pd.IntervalIndex.from_arrays(df1.start_date, df1.end_date, closed='both')
df1['holidays'] = np.sum([iix.contains(x) for x in df2.holiday], axis=0) >= 1
Out[812]:
start_date end_date qty holidays
1 2018-01-01 2018-01-08 23 True
2 2018-01-08 2018-01-15 21 False
3 2018-01-15 2018-01-22 5 False
4 2018-01-22 2018-01-29 12 True
Note: I assume start_date, end_date, holiday columns are in datetime format. If they are not, you need to convert them before run above command as follows
df1.start_date = pd.to_datetime(df1.start_date)
df1.end_date = pd.to_datetime(df1.end_date)
df2.holiday = pd.to_datetime(df2.holiday)

pandas: selecting rows in a specific time window

I have a dataset of samples covering multiple days, all with a timestamp.
I want to select rows within a specific time window. E.g. all rows that were generated between 1pm and 3 pm every day.
This is a sample of my data in a pandas dataframe:
22 22 2018-04-12T20:14:23Z 2018-04-12T21:14:23Z 0 6370.1
23 23 2018-04-12T21:14:23Z 2018-04-12T21:14:23Z 0 6368.8
24 24 2018-04-12T22:14:22Z 2018-04-13T01:14:23Z 0 6367.4
25 25 2018-04-12T23:14:22Z 2018-04-13T01:14:23Z 0 6365.8
26 26 2018-04-13T00:14:22Z 2018-04-13T01:14:23Z 0 6364.4
27 27 2018-04-13T01:14:22Z 2018-04-13T01:14:23Z 0 6362.7
28 28 2018-04-13T02:14:22Z 2018-04-13T05:14:22Z 0 6361.0
29 29 2018-04-13T03:14:22Z 2018-04-13T05:14:22Z 0 6359.3
.. ... ... ... ... ...
562 562 2018-05-05T08:13:21Z 2018-05-05T09:13:21Z 0 6300.9
563 563 2018-05-05T09:13:21Z 2018-05-05T09:13:21Z 0 6300.7
564 564 2018-05-05T10:13:14Z 2018-05-05T13:13:14Z 0 6300.2
565 565 2018-05-05T11:13:14Z 2018-05-05T13:13:14Z 0 6299.9
566 566 2018-05-05T12:13:14Z 2018-05-05T13:13:14Z 0 6299.6
How do I achieve that? I need to ignore the date and just evaluate the time component. I could traverse the dataframe in a loop and evaluate the date time in that way, but there must be a more simple way to do that..
I converted the messageDate which was read a a string to a dateTime by
df["messageDate"]=pd.to_datetime(df["messageDate"])
But after that I got stuck on how to filter on time only.
Any input appreciated.

datetime columns have DatetimeProperties object, from which you can extract datetime.time and filter on it:
import datetime
df = pd.DataFrame(
[
'2018-04-12T12:00:00Z', '2018-04-12T14:00:00Z','2018-04-12T20:00:00Z',
'2018-04-13T12:00:00Z', '2018-04-13T14:00:00Z', '2018-04-13T20:00:00Z'
],
columns=['messageDate']
)
df
messageDate
# 0 2018-04-12 12:00:00
# 1 2018-04-12 14:00:00
# 2 2018-04-12 20:00:00
# 3 2018-04-13 12:00:00
# 4 2018-04-13 14:00:00
# 5 2018-04-13 20:00:00
df["messageDate"] = pd.to_datetime(df["messageDate"])
time_mask = (df['messageDate'].dt.hour >= 13) & \
(df['messageDate'].dt.hour <= 15)
df[time_mask]
# messageDate
# 1 2018-04-12 14:00:00
# 4 2018-04-13 14:00:00

I hope the code is self explanatory. You can always ask questions.
import pandas as pd
# Prepping data for example
dates = pd.date_range('1/1/2018', periods=7, freq='H')
data = {'A' : range(7)}
df = pd.DataFrame(index = dates, data = data)
print df
# A
# 2018-01-01 00:00:00 0
# 2018-01-01 01:00:00 1
# 2018-01-01 02:00:00 2
# 2018-01-01 03:00:00 3
# 2018-01-01 04:00:00 4
# 2018-01-01 05:00:00 5
# 2018-01-01 06:00:00 6
# Creating a mask to filter the value we with to have or not.
# Here, we use df.index because the index is our datetime.
# If the datetime is a column, you can always say df['column_name']
mask = (df.index > '2018-1-1 01:00:00') & (df.index < '2018-1-1 05:00:00')
print mask
# [False False True True True False False]
df_with_good_dates = df.loc[mask]
print df_with_good_dates
# A
# 2018-01-01 02:00:00 2
# 2018-01-01 03:00:00 3
# 2018-01-01 04:00:00 4

df=df[(df["messageDate"].apply(lambda x : x.hour)>13) & (df["messageDate"].apply(lambda x : x.hour)<15)]
You can use x.minute, x.second similarly.

try this after ensuring messageDate is indeed datetime format as you have done
df.set_index('messageDate',inplace=True)
choseInd = [ind for ind in df.index if (ind.hour>=13)&(ind.hour<=15)]
df_select = df.loc[choseInd]
you can do the same, even without making the datetime column as an index, as the answer with apply: lambda shows
it just makes your dataframe 'better looking' if the datetime is your index rather than numerical one.

finding time slots present between start time and end time in python

We have csv file containing predefined time slots.
According to start time and end time provided by the user we want time slots present between the start time and end time.
eg
start time =11:00:00
end time=19:00:00
output- slot_no 2,3,4,5

I think you need boolean indexing with loc and between for selecting column Slot_no, all columns and values are converted to_timedelta, also midnight is replaced to 24:00:00:
df = pd.DataFrame(
{'Slot_no':[1,2,3,4,5,6,7],
'start_time':['0:01:00','8:01:00','10:01:01','12:01:00','14:01:00','18:01:01','20:01:00'],
'end_time':['8:00:00','10:00:00','12:00:00','14:00:00','18:00:00','20:00:00','0:00:00']})
df = df.reindex_axis(['Slot_no','start_time','end_time'], axis=1)
df['start_time'] = pd.to_timedelta(df['start_time'])
df['end_time'] = pd.to_timedelta(df['end_time'].replace('0:00:00', '24:00:00'))
print (df)
Slot_no start_time end_time
0 1 00:01:00 0 days 08:00:00
1 2 08:01:00 0 days 10:00:00
2 3 10:01:01 0 days 12:00:00
3 4 12:01:00 0 days 14:00:00
4 5 14:01:00 0 days 18:00:00
5 6 18:01:01 0 days 20:00:00
6 7 20:01:00 1 days 00:00:00
start = pd.to_timedelta('11:00:00')
end = pd.to_timedelta('19:00:00')
mask = df['start_time'].between(start, end) | df['end_time'].between(start, end)
s = df.loc[mask, 'Slot_no']
print (s)
2 3
3 4
4 5
5 6
Name: Slot_no, dtype: int64
L = df.loc[mask, 'Slot_no'].tolist()
print (L)
[3, 4, 5, 6]