I have numpy array agent which contains 'y' or 'n'. I wanted to replace 'y' with 1 and 'n' with 0 and where something else is present say nan I want to assign -1. Script I wrote was
agent[agent=='y']=1
agent[agent=='n']=0
agent[(agent!='y') and (agent!='n')]=-1
agent=agent.astype(int)
It gave error "The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()" I understand I can use loops but I want to do this in one line in as simple way as I can
Why not first make an array of -1, then fill with 1 and 0 based on agent
result = np.ones_like(agent, dtype=np.int) *-1
result [agent == 'y'] = 1
result [agent == 'n'] = 0
First of all, given the order of assignments you have, if it worked, you'd have your whole array filled with -1's.
That being said, you can do:
agent[(agent != 0) & (agent != 1)] = -1
You can also consider using a masked array.
I would recommend creating a transformation function, f, that returns 0 for 'n', 1 for 'y', and -1 otherwise. Then I would do: desired_array = [f(cell) for cell in agent]
2 * (agent == 'y').astype('int') + (agent == 'n').astype('int') - 1works, but probably not a good idea :-)