I feel kinda stupid as I think this must be easier than I am making it. Below my question you will see an answer I got to a question that I thought I would be able to complete the last steps my self I was wrong :)

 So if I have an array

Y = np.rec.array([(1.0, 0.0, 3.0, 3.5), (0.0, 0.0, 6.0, 6.5), (1.0, 1.0, 9.0, 9.5)], dtype=[('var1', '<f8'), ('var2', '<f8'), ('var3', '<f8'), ('var4', '<f8')])

do this works like I would expect

Y[['var3','var4']][Y['var1']==1]

>>>array([(3.0, 3.5), (9.0, 9.5)], 

      dtype=[('var3', '<f8'), ('var4', '<f8')])

But I would like to do this, 

>>> Y[['var3','var4']][Y['var1']==0 and Y['var2']==0]

Traceback (most recent call last):

  File "<string>", line 1, in <fragment>

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

I tried some any() and all() combination but nothing worked. What s the right way to go about this?

Thanks

Vincent

This answer was received on the mail list from Skipper Seabold

If you have a rec array

Y = np.rec.array([(1.0, 2.0, 3.0), (4.0, 5.0, 6.0), (7.0, 8.0, 9.0)],

dtype=[('var1', ''var2','var3'])

You can access the rows like,

Y[['var1','var2','var3']]

Note the list within [].

If you want a "normal" array, I like this way that Pierre recently

pointed out. 3 is the number of columns, and it fills in the number

of rows.

Y[['var1','var2','var3']].view((float,3))

note the tuple for the view, if they're all floats. Taking a view

might not work if var# have different types, like ints and floats.

If you want the mean of the rows (mean over the columns axis = 1)

Y[['var1','var2','var3']].view((float,3)).mean(1)

Some shortcuts.

Y[list(Y.dtype.names)].view((float,len(Y.dtype))).mean(1)

Also, for now, the columns will given back to you in the order they're

in in the array no matter which way you ask for them. A patch has

been submitted for returning the order you ask that I hope gets picked

up...