Proper pandas map syntax with multiple values

Question

I know I'm almost there, but can't figure out the faster syntax for the equivalent of the below code:

def is_evening(df):
    lst = []
    for i in df.index:
        val = df["localized_time"][i]
        if val != 'NA':
            if val.hour >= 17 and val.hour <= 3:
                lst.append(1)
            else:
                lst.append(0)
        else:
            lst.append(-1)
    df["is_evening"] = lst
    return df

If it was just two values it would be easy, but with the additional conditions and stuff, is there a fast way to do this using pandas map?

What I have so far:

mask = [df["localized_time"] != 'NA']
df["is_evening"] = df["is_evening"] = df["localized_time"][mask].apply(lambda x: 1 if (x.hour >= 17 and x.hour <= 23) else 0)
# Doesn't seem to work

Could anyone help me out? Thank you! :)

I can't imagine when val.hour >= 17 and val.hour <= 3 would be True ;-) — MaxU - stand with Ukraine
– MaxU - stand with Ukraine, Commented May 20, 2017 at 18:06
@MaxU Holy crap tired coding does not help, thank you for spotting that! — Wboy
– Wboy, Commented May 21, 2017 at 3:45

willeM_ Van Onsem · Accepted Answer · 2017-05-20 16:25:38Z

1

Yes there definitely is:

def f(x):
    return int(x.hour >= 17 and x.hour <= 3) if x != 'NA' else -1

df["is_evening"] = df["localized_time"].apply(f)

So here f is a function that maps a single localized_time element to an is_evening element. We can also use a lambda expression, but this will be more ugly:

df["is_evening"] = df["localized_time"].\
    apply(lambda x: int(x.hour >= 17 and x.hour <= 3) if x != 'NA' else -1)

edited May 20, 2017 at 16:25

answered May 20, 2017 at 16:07

willeM_ Van Onsem

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2 Comments

Wboy Over a year ago

Awesome! Thank you so much! :)

DSM Over a year ago

Note that while this is more concise, it still uses Python-level functions instead of vectorized operations.

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