Change dtype for particular values in numpy array?

Question

I have a numpy array x, dimensions = (20, 4), in which only the first row and column are real string values (alphabets) and rest of the values are numerals with their types allocated as string. I want to change these numeral values to float or integer type.

I have tried some steps:

a. I made copies of first row and column of the array as separate variables:

x_row = x[0]
x_col = x[:,0]

Then deleted them from the original array x (using numpy.delete() method) and convertd the type of remaining values by applying a for loop that iterates over each value. However, when I stack back the copied rows and columns using numpy.vstack() and numpy.hstack(), then everything again converts to strings type. So, not sure why this is happening.

b. Same procedure as point a, except I used numpy.insert() method for inserting rows and columns, but is doing the same thing - converting everything back to string type.

So, is there a way through which I don't have to go through this deleting and stacking mechanism (which isn't working anyways) and I can change all the values (except first row and column) of an array to int() or float() type?

BrenBarn · Accepted Answer · 2013-01-03 01:04:50Z

4

All items in a numpy array have to have the same dtype. That is a fundamental fact about numpy. You could possibly use a numpy recarray, or you could use dtype=object which basically lets all values be anything.

I'd recommend you take a look at pandas, which provides a tabular data structure that allows different columns to have different types. It sounds like what you have is a table with row and column labels, and that's what pandas deals with nicely.

edited Jan 3, 2013 at 1:04

answered Jan 2, 2013 at 22:58

BrenBarn

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1 Comment

saifuddin778 Over a year ago

You are absolutely right friend..haven't noticed that. I think rec array or dtype does my job (i.e. allocating the dtype of the array as object (dtype=object))... Also, never tried pandas arrays but read about them a bit. Thanks BrenBarn. :-)

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