IF else and for loop in one line

Question

I need to apply if else condition and for loop in single line.I need to update both 'RL' and "RM" at a time and update other values as 'Others'.How to do it??.IS it possible??

train['MSZoning']=['RL' if x=='RL' else 'Others' for x in train['MSZoning']]

@Anesh - Sure, both solutions works, but better is avoid loops, avoid `apply` because slow. Check it in my timings in edited answer. — jezrael, Mar 30 '18 at 13:08
Yes I saw it.Can you explain your code.What does `isin` does?? — Anesh, Mar 30 '18 at 15:09

jezrael · Accepted Answer · 2018-03-30T13:07:42.110

Use numpy.where:

train['MSZoning'] = np.where(train['MSZoning'] == 'RM', 'RM', 'Others')

If need update all without RM and RL use isin with inverted boolean mask by ~:

train = pd.DataFrame({'MSZoning':['RL'] *3 + ['qa','RM','as']})
train.loc[~train['MSZoning'].isin(['RM','RL']), 'MSZoning'] =  'Others'

print (train)
  MSZoning
0       RL
1       RL
2       RL
3   Others
4       RM
5   Others

Timings:

train = pd.DataFrame({'MSZoning':['RL'] *3 + ['qa','RM','as']})
#[60000 rows x 1 columns]
train = pd.concat([train] * 10000, ignore_index=True)

In [202]: %timeit train.loc[~train['MSZoning'].isin(['RM','RL']), 'MSZoning'] =  'Others'
5.82 ms ± 447 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [203]: %timeit train['MSZoning'] = train['MSZoning'].apply(lambda x: x if x in ('RM', 'RL') else 'Others')
15 ms ± 584 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

score 1 · Answer 2 · answered Mar 30 '18 at 12:41

1

So if you want to keep RM and RL while marking others as Others you can use:

train['MSZoning'] = train['MSZoning'].apply(lambda x: x if x in ('RM', 'RL') else 'Others')

answered Mar 30 '18 at 12:41

zipa

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IF else and for loop in one line

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