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returns the coordinates of a numpy array given the index and the shape. A first_index_et function is given as example code

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# -*- coding: utf-8 -*-
"""
Created on Fri Oct 05 13:32:06 2012

@author: Garrett Berg
"""
import numpy as np
import itertools
import math

def index_to_coords(index, shape):
    '''convert index to coordinates given the shape'''
    coords = []
    for i in xrange(1, len(shape)):
        divisor = int(np.product(shape[i:]))
        value = index//divisor
        coords.append(value)
        index -= value * divisor
    coords.append(index)
    return tuple(coords)
    
def first_coords_et(data_matrix, value, start = 0):
    '''the first coordinates that are equal to the value'''
    index = first_index_et(data_matrix.flatten(), value, start)
    shape = data_matrix.shape
    return index_to_coords(index, shape)

def first_index_et(data_list, value, start = 0):
    data_list = itertools.islice(data_list, start, None)
    '''same as data_list.index(value), except with exception handling
    Also finds 'nan' values and works with numpy arrays -- quickly!'''
    try:
        if type(value) == float and math.isnan(value):
            floats = (float, np.float64, np.float32, np.float96)
            isnan = math.isnan
            return next(data[0] for data in enumerate(data_list)
              if (type(data[1]) in floats
              and isnan(data[1])))  + start
        else:
            return next(data[0] for data in 
            enumerate(data_list) if data[1] == value) + start
    except (ValueError, StopIteration): return - 1

if __name__ == '__main__':
    ab = np.arange(0, 30000)
    ab.shape = (20, 20, 30000/(20*20))
    value = ab[7][12][0]
    print first_coords_et(ab, value)
    # (7, 12, 0)