Try:

myData.dtype.names

This will return a tuple of the field names.

In [10]: myData.dtype.names
Out[10]: ('TIME', 'FX', 'FY', 'FZ')

NumPy structured arrays have named columns:

import numpy as np
    
a = range(100)
A = np.array(list(zip(*[iter(a)] * 2)), dtype=[('C1', 'int32'),('C2', 'int64')])
print(A.dtype)

[('C1', '<i4'), ('C2', '<i8')]

You can access the columns by name like this:

print(A['C1'])
# [ 0  2  4  6  8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48
#  50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98]

Note that using np.array with zip causes NumPy to build an array from a temporary list of tuples. Python lists of tuples use a lot more memory than equivalent NumPy arrays. So if your array is very large you may not want to use zip.

Instead, given a NumPy array A, you could use ravel() to make A a 1D array, and then use view to turn it into a structured array, and then use astype to convert the columns to the desired type:

a = range(100)
A = np.array(a).reshape( len(a)//2, 2)
A = A.ravel().view([('col1','i8'),('col2','i8'),]).astype([('col1','i4'),('col2','i8'),])
print(A[:5])
# array([(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)], 
#       dtype=[('col1', '<i4'), ('col2', '<i8')])

print(A.dtype)
# dtype([('col1', '<i4'), ('col2', '<i8')])

The problem is that you are thinking in terms of spreadsheet-like arrays, whereas NumPy does use different concepts.

Here is what you must know about NumPy:

1. NumPy arrays only contain elements of a single type.
2. If you need spreadsheet-like "columns", this type must be some tuple-like type. Such arrays are called Structured Arrays, because their elements are structures (i.e. tuples).

In your case, NumPy would thus take your 2-dimensional regular array and produce a one-dimensional array whose type is a 108-element tuple (the spreadsheet array that you are thinking of is 2-dimensional).

These choices were probably made for efficiency reasons: all the elements of an array have the same type and therefore have the same size: they can be accessed, at a low-level, very simply and quickly.

Now, as user545424 showed, there is a simple NumPy answer to what you want to do (genfromtxt() accepts a names argument with column names).

If you want to convert your array from a regular NumPy ndarray to a structured array, you can do:

data.view(dtype=[(n, 'float64') for n in csv_names]).reshape(len(data))

(you were close: you used astype() instead of view()).

You can also check the answers to quite a few Stackoverflow questions, including Converting a 2D numpy array to a structured array and how to convert regular numpy array to record array?.

First because your question asks about giving names to arrays, I feel obligated to point out that using "structured arrays" for the purpose of giving names is probably not the best approach. We often like to give names to rows/columns when we're working with tables, if this is the case I suggest you try something like pandas which is awesome. If you simply want to organize some data in your code, a dictionary of arrays is often much better than a structured array, so for example you can do:

Y = {'ID':X[0], 'Ring':X[1]}

With that out of the way, if you want to use a structured array, here is the clearest way to do it in my opinion:

import numpy as np

RING = [1,2,2,3,3,3]
ID = [1,2,3,4,5,6]
X = np.array([ID, RING])

dt = {'names':['ID', 'Ring'], 'formats':[int, int]}
Y = np.zeros(len(RING), dtype=dt)
Y['ID'] = X[0]
Y['Ring'] = X[1]

The correct data input form for a structured array is a list of tuples:

In [71]: signal = [(1,2,3),(2,3,1),(3,2,1)] 
    ...: col_names = ('left','right','center') 
    ...: signal = np.array(signal, dtype = [(n, 'int16') for n in col_names])   
In [72]:                                                                        
In [72]: signal                                                                 
Out[72]: 
array([(1, 2, 3), (2, 3, 1), (3, 2, 1)],
      dtype=[('left', '<i2'), ('right', '<i2'), ('center', '<i2')])

1.16 has added a couple of functions that make it easier to convert to and from structured arrays:

In [73]: import numpy.lib.recfunctions as rfn                                   
In [74]: signal = np.array([[1,2,3],[1,2,3],[1,2,3]])                           
In [75]: dt = np.dtype([(n, 'int16') for n in col_names])                       
In [76]: dt                                                                     
Out[76]: dtype([('left', '<i2'), ('right', '<i2'), ('center', '<i2')])
In [77]: rfn.unstructured_to_structured(signal, dt)                             
Out[77]: 
array([(1, 2, 3), (1, 2, 3), (1, 2, 3)],
      dtype=[('left', '<i2'), ('right', '<i2'), ('center', '<i2')])

Applying this dt to signal has a problem:

In [82]: signal.view(dt)                                                        
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-82-f0fa01ce8128> in <module>
----> 1 signal.view(dt)

ValueError: When changing to a smaller dtype, its size must be a divisor of the size of original dtype

We can get around that by first converting signal to a compatible dtype:

In [83]: signal.astype('i2').view(dt)                                           
Out[83]: 
array([[(1, 2, 3)],
       [(1, 2, 3)],
       [(1, 2, 3)]],
      dtype=[('left', '<i2'), ('right', '<i2'), ('center', '<i2')])

But note that Out[83] shape is (3,1). The other arrays were shape (3,). view has always had this shape problem when converting to/from structured arrays. That's part of why the newer functions are easier to use.

As far as I know it's not possible to "name" the rows with pure structured NumPy arrays.

But if you have pandas it's possible to provide an "index" (which essentially acts like a "row name"):

>>> import pandas as pd
>>> import numpy as np
>>> column_names = ['a', 'b', 'c']
>>> row_names    = ['1', '2', '3']

>>> matrix = np.reshape((1, 2, 3, 4, 5, 6, 7, 8, 9), (3, 3))
>>> df = pd.DataFrame(matrix, columns=column_names, index=row_names)
>>> df
   a  b  c
1  1  2  3
2  4  5  6
3  7  8  9

>>> df['a']['3']      # first "column" then "row"
7

>>> df.loc['3', 'a']  # another way to index "row" and "column"
7

Use the following code:

import re

f = open('f.csv','r')

alllines = f.readlines()
columns = re.sub(' +',' ',alllines[0]) #delete extra space in one line
columns = columns.strip().split(',') #split using space

print(columns)

Assume CSV file is like this:

xy   wz  hi kq
0    10  5  6
1    2   4  7
2    5   2  6