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

user_15488802 0
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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)
python
5 Answers

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)
user_7708300

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
user_79032430
user_64172900 
dict = {
  "Monday": 84,
  "Tuesday": 96,
  ...
}

You can make use of dictionary here

user_170653090

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)
user_122118200
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