Python. IF condition using Lambda function in Pandas dataframe

Question

df = pd.read_csv('data/eurusd_dukascopy.csv')
df.columns = ['timestamp', 'open', 'high', 'low', 'close', 'volume']
df['oc'] = df.close - df.open

df['uptail'] = df['oc'].apply(lambda x: (df.high - df.close) if x >= 0 else (df.high - df.open))

Gives an error: ValueError: Wrong number of items passed 2963, placement implies 1

I just want to do the following: if df.oc is a positive number, then df.uptail = (df.high - df.close) ...else df.uptail = (df.high - df.open)

How can I crank this out?

I tried on a phone but got lost in the logic. You probably want np.where but it's not clear exactly how it works for your example. — roganjosh
– roganjosh, Commented Mar 8, 2019 at 17:47

mad_ · Accepted Answer · 2019-03-08 17:58:57Z

1

np.where is more suited here.

df['uptail'] = np.where(df.close-df.open>=0, df.high-df.close, df.high-df.open)

edited Mar 8, 2019 at 17:58

answered Mar 8, 2019 at 17:52

mad_

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1 Comment

Swier Over a year ago

For brevity: df.close >= df.open equals df.close - df.open >= 0

Swier · Accepted Answer · 2019-03-08 17:52:38Z

0

It looks like you want uptail to be close - open if that's positive, and high - open if close - open is negative or 0.

You can achieve that with the following code:

df['uptail'] = df.high - df.open
df.loc[df.close > df.open, 'uptail'] = \
    df.loc[df.close > df.open, 'high'] - df.loc[df.close > df.open, 'close']

answered Mar 8, 2019 at 17:52

Swier

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