np.where() or another boolean way for new pandas dataframe column

Question

I would like to create a column in a dataframe with stock prices that is conditional on the boolean values from another calculation.

              close    high   low
    Index
      0        10       11     10
      1        11       12     10
      2        10       11      9

First I want to define some condition(s) that could logically look like that but in fact are a lengthy :

 condition1: if df.close > df.close.shift() return True

In real I want to define many more conditions that all deliver True or False. Then I include that in the np.where():

 df['NewColumn'] = np.where(condition1() == True, 'A', 'B')

I tried to define the condition as a function but did not manage to correctly set it up. I would like to avoid to write the content of the condition directly into the np.where() because it would become too complex with several nested conditions.

So, how can I accomplish my task most efficiently?

Edit: a function could look like this (but it does not work in the np.where() above):

def condition1(): 
    if df.Close > df.Close.shift(1):
        return True
    Else
        return False

`numpy.where(df.close > df.close.shift(), 'A', 'B')` should work — Paul H, Oct 20 '16 at 20:29
Care to share ***the condition you tried to define as a function but did not manage to correctly set it up***? — Abdou, Oct 20 '16 at 20:30
@Paul H, I want to avoid writing the condition directly into the formula due to the complexity and number of the real conditions — Al_Iskander, Oct 20 '16 at 20:37

score 2 · Accepted Answer · edited May 23 '17 at 10:32

UPDATE: IMO you don't need "conditions" as functions:

In [89]: cond1 = df.Close > df.Close.shift(1)

In [90]: cond2 = df.High < 12

In [91]: df['New_Close'] = np.where(cond1, 'A', 'B')

In [92]: df
Out[92]:
   Close  High  Low New_Close
0     10    11   10         B
1     11    12   10         A
2     10    11    9         B

In [93]: df['New_High'] = np.where(cond1, '<12', '>=12')

In [94]: df
Out[94]:
   Close  High  Low New_Close New_High
0     10    11   10         B     >=12
1     11    12   10         A      <12
2     10    11    9         B     >=12

OLD answer:

I don't really see any advantages (benefit) of this approach, but you can do it this way:

def condition1(): 
    return (df.Close > df.Close.shift(1))

def condition2(): 
    return (df.High < 12)


In [72]: df['New_Close'] = np.where(condition1(), 'A', 'B')

In [73]: df
Out[73]:
   Close  High  Low New_Close
0     10    11   10         B
1     11    12   10         A
2     10    11    9         B

In [74]: df['New_High'] = np.where(condition2(), '<12', '>=12')

In [75]: df
Out[75]:
   Close  High  Low New_Close New_High
0     10    11   10         B      <12
1     11    12   10         A     >=12
2     10    11    9         B      <12

PS IMO it would be easier and nicer to do it directly like @PaulH said in his comment

thank you, this seems to work. The benefit is mostly optical in nature. Should improve the readability of the np.where() when it becomes long and nested. — Al_Iskander, Oct 20 '16 at 20:47

np.where() or another boolean way for new pandas dataframe column

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