Filling a dataframe column based on multiple conditional choices

Question

I have a dataframe data which looks like:

Holiday  Report   Action     Power
0             0        0     1.345
0             0        0     1.345
1             0        0     0
0             0        0     1.345
0             0        0     1.345
0             1        0     0
0             0        0     1.345
0             1        1     0
0             0        0     1.345
0             0        1     0

Columns 'Holiday', 'Report' and 'Actions' are pulled in. Power is calculated by looking a some other columns of data. I am looking for a line of code that will allow me to set the power to zero if any of the columns 'Holiday', 'Report' or 'Actions' is set to 1.

I can do this procedure one column at a time:

data.loc[staticidx, "power"] = np.where(dayData.Holiday = 1, 0, (a - b)/(c-1))
data.loc[staticidx, "power"] = np.where(dayData.Report= 1, 0, (a - b)/(c-1))
data.loc[staticidx, "power"] = np.where(dayData.Actions= 1, 0, (a - b)/(c-1))

but is there a way to combine them into one function with something like an OR statement?

Many thanks

Answer 1

You can subselect the rows by slicing, compare against 1 and use any with param axis=1 and create a boolean mask passing to loc to set only those rows that meet the condition to 1 as desired:

In [23]:
df.loc[(df.ix[:,:'Action'] == 1).any(axis=1), 'Power'] = 1
df

Out[23]:
   Holiday  Report  Action  Power
0        0       0       0  1.345
1        0       0       0  1.345
2        1       0       0  1.000
3        0       0       0  1.345
4        0       0       0  1.345
5        0       1       0  1.000
6        0       0       0  1.345
7        0       1       1  1.000
8        0       0       0  1.345
9        0       0       1  1.000

Answer 2

Hope this helps using np.where :

In[49]:df['Power']=np.where((df['Holiday']==1)|(df['Report']==1)|(df['Action']==1),1,df['Power'])

In[50]:df
Out[50]: 
   Holiday  Report  Action  Power
0        0       0       0  1.345
1        0       0       0  1.345
2        1       0       0  1.000
3        0       0       0  1.345
4        0       0       0  1.345
5        0       1       0  1.000
6        0       0       0  1.345
7        0       1       1  1.000
8        0       0       0  1.345
9        0       0       1  1.000