How to simplify python code that has many if-else statements?

Question

so I have this code but I don't want to use many if-else condition and I wonder if I could simplify this. Any idea for this? Can I use a loop for this?

    if dh1['date1'][0].strftime("%A") == 'Monday':
        df=df=pd.concat([dh1,dh2.tail(84)])
        df=df.sort_values(['date1','hr1'])
        df=df.reset_index()
        df=df.drop('index', 1)
    elif dh1['date1'][0].strftime("%A") == 'Tuesday':
        df=df=pd.concat([dh1,dh2.tail(96)])
        df=df.sort_values(['date1','hr1'])
        df=df.reset_index()
        df=df.drop('index', 1)
    elif dh1['date1'][0].strftime("%A") == 'Wednesday':
        df=df=pd.concat([dh1,dh2.tail(108)])
        df=df.sort_values(['date1','hr1'])
        df=df.reset_index()
        df=df.drop('index', 1)
    elif dh1['date1'][0].strftime("%A") == 'Thursday':
        df=df=pd.concat([dh1,dh2.tail(120)])
        df=df.sort_values(['date1','hr1'])
        df=df.reset_index()
        df=df.drop('index', 1)
    elif dh1['date1'][0].strftime("%A") == 'Friday':
        df=df=pd.concat([dh1,dh2.tail(132)])
        df=df.sort_values(['date1','hr1'])
        df=df.reset_index()
        df=df.drop('index', 1)
    elif dh1['date1'][0].strftime("%A") == 'Saturday':
        df=df=pd.concat([dh1,dh2.tail(144)])
        df=df.sort_values(['date1','hr1'])
        df=df.reset_index()
        df=df.drop('index', 1)
    elif dh1['date1'][0].strftime("%A") == 'Sunday':
        df=df=pd.concat([dh1,dh2.tail(156)])
        df=df.sort_values(['date1','hr1'])
        df=df.reset_index()
        df=df.drop('index', 1)

optimize your code and use match case

Mehmaam
– Mehmaam

2022-08-18 06:01:49 +00:00
Commented Aug 18, 2022 at 6:01 — Mehmaam
– Mehmaam, Commented Aug 18, 2022 at 6:01

bereal · Accepted Answer · 2022-08-18 06:02:59Z

2

It seems that the only difference between all the branches is the argument passed to the tail method. Moreover, the difference between the argument value for the adjacent days is 12, so it can be evaluated as 84 + 12 * weekday counting from Monday as 0. If that's really the case, you can reduce the code like this:

arg = 84 + dh1['date'][0].weekday() * 12
df=df=pd.concat([dh1,dh2.tail(arg)])
df=df.sort_values(['date1','hr1'])
df=df.reset_index()
df=df.drop('index', 1)

answered Aug 18, 2022 at 6:02

bereal

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Comments

shivankgtm · Accepted Answer · 2022-08-18 06:03:28Z

2

The only difference is of df.tail parameter, which depends on date1 column. whenever these cases come you create a mapping like following.

tail_day_map = {
    'Monday': 84,
    'Tuesday': 96,
    'Wednesday': 108,
    'Thursday': 120,
    'Friday': 132,
    'Saturday': 144,
    'Sunday': 156
}

def perform_action(df, tail_day_map):
    tail_number = tail_day_map[df['date1'][0].strftime("%A")]
    df = df.tail(tail_number)
    df = df.sort_values(['date1','hr1'])
    df = df.reset_index()
    df = df.drop('index', 1)
    return df

answered Aug 18, 2022 at 6:03

shivankgtm

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3 Comments

Michael Pruglo Over a year ago

instead of map, you can simplify it even further to 84 + 12 * weekday_int

bereal Over a year ago

One problem with this code (as well as with the original one) is that %A returns a locale-specific day name, so it will break if it runs on any computer with non-English language.

shivankgtm Over a year ago

@bereal thanks for the point, i was not aware of this point.

GeekPete · Accepted Answer · 2022-08-18 06:09:19Z

1

If you're able to use Python 3.10 or later, how about using Match Case? https://learnpython.com/blog/python-match-case-statement/

Match Case may simplify your code and provide more readability, but as mentioned in the second link be cautious with the order of the cases as it may change the behavior of the logic.

Otherwise, there may be other flow control strategies you can use: https://docs.python.org/3/tutorial/controlflow.html

answered Aug 18, 2022 at 6:09

GeekPete

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Comments

Rickfdalton · Accepted Answer · 2022-08-18 06:02:52Z

0

dict = {
  "Monday": 84,
  "Tuesday": 96,
  ...
}

You can make use of dictionary here

answered Aug 18, 2022 at 6:02

Rickfdalton

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Comments

CNFranc11s · Accepted Answer · 2022-08-18 06:07:10Z

0

first,
df=df=pd.concat([dh1,dh2.tail(156)]) maybe can be
df=pd.concat([dh1,dh2.tail(156)])
then,

df=df.sort_values(['date1','hr1'])
df=df.reset_index()
df=df.drop('index', 1)

can put out of the if condition.
then for me,
I'll make a dict like: d = {"Monday":84,"Tuesday":96...} on the out of if condition use:

weekday = dh1['date1'][0].strftime("%A")
df=pd.concat([dh1,dh2.tail(d[weekday])])

so the final code like:

d = {"Monday":84,"Tuesday":96...}
weekday = dh1['date1'][0].strftime("%A")
df=pd.concat([dh1,dh2.tail(d[weekday])])
df=df.sort_values(['date1','hr1'])
df=df.reset_index()
df=df.drop('index', 1)

answered Aug 18, 2022 at 6:07

CNFranc11s

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