How to iterate over hours of a given day in python?

Question

I have the following time series data of temperature readings:

DT                Temperature
01/01/2019 0:00     41
01/01/2019 1:00     42
01/01/2019 2:00     44
......
01/01/2019 23:00    41
01/02/2019 0:00     44

I am trying to write a function that compares the hourly change in temperature for a given day. Any change greater than 3 will increment quickChange counter. Something like this:

def countChange(day):
    for dt in day:
        if dt+1 - dt > 3: quickChange = quickChange+1

I can call the function for a day ex: countChange(df.loc['2018-01-01'])

Can you create [minimal, complete, and verifiable example](http://stackoverflow.com/help/mcve)? — jezrael, Dec 04 '19 at 06:32

jezrael · Accepted Answer · 2019-12-04T06:56:26.440

Use Series.diff with compare by 3 and count Trues values by sum:

np.random.seed(2019)

rng = (pd.date_range('2018-01-01', periods=10, freq='H').tolist() +
      pd.date_range('2018-01-02', periods=10, freq='H').tolist())
df = pd.DataFrame({'Temperature': np.random.randint(100, size=20)}, index=rng)  
print (df)
                     Temperature
2018-01-01 00:00:00           72
2018-01-01 01:00:00           31
2018-01-01 02:00:00           37
2018-01-01 03:00:00           88
2018-01-01 04:00:00           62
2018-01-01 05:00:00           24
2018-01-01 06:00:00           29
2018-01-01 07:00:00           15
2018-01-01 08:00:00           12
2018-01-01 09:00:00           16
2018-01-02 00:00:00           48
2018-01-02 01:00:00           71
2018-01-02 02:00:00           83
2018-01-02 03:00:00           12
2018-01-02 04:00:00           80
2018-01-02 05:00:00           50
2018-01-02 06:00:00           95
2018-01-02 07:00:00            5
2018-01-02 08:00:00           24
2018-01-02 09:00:00           28

#if necessary create DatetimeIndex if DT is column
df = df.set_index("DT")

def countChange(day):
    return (day['Temperature'].diff() > 3).sum()

print (countChange(df.loc['2018-01-01']))
4

print (countChange(df.loc['2018-01-02']))
9

Thanks, since it is comparing with the previous hours, I would also like to consider the negative change. So +3 and -3, how can I incorporate that? — Sakib Shahriar, Dec 04 '19 at 06:59
@SakibShahriar - then change `(day['Temperature'].diff() > 3).sum()` to `(day['Temperature'].diff().abs() > 3).sum()` — jezrael, Dec 04 '19 at 07:00

Iron Hand Odin · Answer 2 · 2019-12-04T06:51:17.307

try pandas.DataFrame.diff:

df = pd.DataFrame({'dt': ["01/01/2019 0:00","01/01/2019 1:00","01/01/2019 2:00","01/01/2019 23:00","01/02/2019 0:00"], 
                    'Temperature': [41, 42, 44, 41, 44]})

df = df.sort_values("dt")
df = df.set_index("dt")

def countChange(df):
    df["diff"] = df["Temperature"].diff()
    return df.loc[df["diff"] > 3, "diff"].count()

quickchange = countChange(df.loc["2018-01-01"])

How to iterate over hours of a given day in python?

2 Answers2