New column based on a row with conditions in Pandas - python-3.x

Question

I'm trying to do an operation with Dataframes but i'm not sure how I can solve the problem using the built-in Pandas Operations (Actualy my code is based on a for so I'm trying to build a more elegant solution).
Given the following Dataframes, defined with the columns described below
original_df = [o1, o2, o3, o4]
weights_df = [w1, w2, w3, w4]
conditions_df = [c1, c2, c3, c4]
I need to built a new column on original_df based on the division of o1/w1 but depending on the value of c1, with takes the values ["+" or "-" I need to do the -o1/w1 operation.
As long as I did was:
orignal_df['newcolumn'] = original_df / weights_df
Where of course I divided the two terms but without applying the condition, I'm trying to do with map and apply functions but I'm not sure how I can add the third column into the function.

Answer 1

original_df = [100, 200, 300, 400]
weights_df = [10, 20, 30, 40]
conditions_df = [1, 2, 3, 4]
df = pd.DataFrame({'x':original_df, 'y':weights_df, 'z':conditions_df})
def div(x, y, z):
if z > 2:
return float(x/y)
else:
return float(-1*x/y)
df['new_feature'] = df.apply(lambda p: div(p['x'], p['y'], p['z']), axis=1)
This is one way of solving. If your conditions_df contains '+'/'-' then you can change the condition in def div(x, y, z) accordingly.

Answer 2

You can use numpy.where for mask by condition:
#data from lisa answer
#df = pd.DataFrame({'x':original_df, 'y':weights_df, 'z':conditions_df})
df['new_feature'] = df['x'] / df['y'] * np.where(df['z'] > 2, 1, -1)
print (df)
x y z new_feature
0 100 10 1 -10.0
1 200 20 2 -10.0
2 300 30 3 10.0
3 400 40 4 10.0
Timings:
#4k rows
df = pd.concat([df]*1000).reset_index(drop=True)
#lisa answer
In [95]: %timeit df['new_feature1'] = df.apply(lambda p: div(p['x'], p['y'], p['z']), axis=1)
10 loops, best of 3: 123 ms per loop
In [96]: %timeit df['new_feature2'] = df['x'] / df['y'] * np.where(df['z'] > 2, 1, -1)
1000 loops, best of 3: 595 µs per loop