Sorting numpy matrix for a given column

Question

I've tried to use Ned Batchelder code to sort in human order a NumPy matrix, as it was proposed in this following post:

Sort numpy string array with negative numbers?

The code runs on a one-dimensional array, the command being:

print (sorted(a, key=natural_keys))

Now, my problem is that my data is a 10 column matrix and I want to sort it according to one column (let's say MyColumn). I can't find a way to modify the code to print the whole matrix sorted according to this very column. All I could come up with is this:

print (sorted(a['MyColumn'], key=natural_keys))

But, of course, only MyColumn shows up in the output, although it is correctly sorted...

Is there a way to print the whole Matrix?

Here is the command I used to load my array (I simplified my original imputfile to a 3 column array):

data = np.loadtxt(inputfile, dtype={'names': ('ID', 'MyColumn', 'length'),
'formats': ('int32', 'S40', 'int32')},skiprows=1, delimiter='\t')

ID  MyColumn    length
164967  BFT_job13_q1_type2  426
197388  BFT_job8_q0_type2   244
164967  BFT_job13_q0_type1  944
72406   BFT_job1_q0_type3   696

Here is what the output would ideally look like:

ID  MyColumn    length
72406   BFT_job1_q0_type3   696
197388  BFT_job8_q0_type2   244
164967  BFT_job13_q0_type1  944
164967  BFT_job13_q1_type2  426

Community · Accepted Answer · 2017-05-23 12:13:23Z

5

If you have a np.matrix, called m:

col = 1
m[np.array(m[:,col].argsort(axis=0).tolist()).ravel()]

If you have a np.ndarray, called a:

col = 1
a[a[:,col].argsort(axis=0)]

If you have a structured array with named columns:

def mysort(data, col_name, key=None):
    d = data.copy()
    cols = [i[0] for i in eval(str(d.dtype))]
    if key:
        argsort = np.array([key(i) for i in d[col_name]]).argsort()
    else:
        argsort = d[col_name].argsort()
    for col in cols:
        d[col] = d[col][argsort]
    return d

For your specific case you need the following key function:

def key(x):
    x = ''.join([i for i in x if i.isdigit() or i=='_'])
    return '{1:{f}{a}10}_{2:{f}{a}10}_{3:{f}{a}10}'.format(*x.split('_'), f='0', a='>')

d = mysort(data, 'MyColumn', key)

edited May 23, 2017 at 12:13

CommunityBot

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answered Jun 30, 2013 at 6:58

Saullo G. P. Castro

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11 Comments

Sara Over a year ago

Thanks. Still can't sort it right... 1-says "IndexError: too many indices" pointing at [:,col] 2- I changed [:,col] to [:][col] and it seems to do the job, but I don't know where to insert key=natural_keys to sort it the right way.

Sara Over a year ago

I might add... I don't know how to check whether I have a np.ndarray or a np.matrix! I loaded a text file with np.loadtxt

Saullo G. P. Castro Over a year ago

you have a np.ndarray when loading with np.loadtxt()

Saullo G. P. Castro Over a year ago

@Sara It should work you don't need natural_keys... you just have to inform which column you want to sort using col, remembering that 0 is the 1st column

Saullo G. P. Castro Over a year ago

for large job numbers just use a higher number of more digits in the formatting string, like '{1:{f}{a}10}_{2:{f}{a}10}_{3:{f}{a}10}'... which can handle a job number up to 9999999999

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