Search for a pattern in a numpy array

Question

Is there an easy way to find all the matching elements in a NumPy array according to some pattern?

For example, consider the following array:

a = array(['zzzz', 'zzzd', 'zzdd', 'zddd', 'dddn', 'ddnz', 'dnzn', 'nznz',
       'znzn', 'nznd', 'zndd', 'nddd', 'ddnn', 'dnnn', 'nnnz', 'nnzn',
       'nznn', 'znnn', 'nnnn', 'nnnd', 'nndd', 'dddz', 'ddzn', 'dznn',
       'znnz', 'nnzz', 'nzzz', 'zzzn', 'zznn', 'dddd', 'dnnd'], dtype=object)

And I need to find all combinations containing "** dd".

I basically need a function that receives an array as input and returns a smaller array with all the relevant elements:

>> b = func(a, pattern='**dd')
>> b = array(['zzdd', 'zddd', 'zndd', 'nddd', 'nndd', 'dddd'], dtype=object)

Answer 1

Since it turns out that you are actually working with pandas, there are easier ways to do this at the Series level instead of just ndarray using vectorized string operations :

In [32]: s = pd.Series(['zzzz', 'zzzd', 'zzdd', 'zddd', 'dddn', 'ddnz', 'dnzn', 'nznz',
    ...:        'znzn', 'nznd', 'zndd', 'nddd', 'ddnn', 'dnnn', 'nnnz', 'nnzn',
    ...:        'nznn', 'znnn', 'nnnn', 'nnnd', 'nndd', 'dddz', 'ddzn', 'dznn',
    ...:        'znnz', 'nnzz', 'nzzz', 'zzzn', 'zznn', 'dddd', 'dnnd'])

In [33]: s[s.str.endswith("dd")]
Out[33]: 
2     zzdd
3     zddd
10    zndd
11    nddd
20    nndd
29    dddd
dtype: object

, ndarray:

In [34]: s[s.str.endswith("dd")].values
Out[34]: array(['zzdd', 'zddd', 'zndd', 'nddd', 'nndd', 'dddd'], dtype=object)

, :

In [49]: s[s.str.match(".*dd$")]
Out[49]: 
2     zzdd
3     zddd
10    zndd
11    nddd
20    nndd
29    dddd
dtype: object

Answer 2

numpy.core.defchararray.rfind, , , 2 . 4 , , 4 - 2 = 2.

, -

a[np.core.defchararray.rfind(a.astype(str),'dd')==2]

---

, , 2, -

len_sub = np.array(list(map(len,a)))-len('dd')
a[np.core.defchararray.rfind(a.astype(str),'dd')==len_sub]

, , dd -

In [121]: a = np.append(a,'ewqjejwqjedd')

In [122]: len_sub = np.array(list(map(len,a)))-len('dd')

In [123]: a[np.core.defchararray.rfind(a.astype(str),'dd')==len_sub]
Out[123]: array(['zzdd', 'zddd', 'zndd', 'nddd', 'nndd', 'dddd',\
                 'ewqjejwqjedd'], dtype=object)

Answer 3

numpy. , numpy, python, python numpy , .

, , , fnmatch ??dd ( 2 + dd )

( re.match() ..dd$ ).

, , take :

from numpy import array
import fnmatch

a = array(['zzzz', 'zzzd', 'zzdd', 'zddd', 'dddn', 'ddnz', 'dnzn', 'nznz',
       'znzn', 'nznd', 'zndd', 'nddd', 'ddnn', 'dnnn', 'nnnz', 'nnzn',
       'nznn', 'znnn', 'nnnn', 'nnnd', 'nndd', 'dddz', 'ddzn', 'dznn',
       'znnz', 'nnzz', 'nzzz', 'zzzn', 'zznn', 'dddd', 'dnnd'], dtype=object)

def func(ar,pattern):
    indices = [i for i,x in enumerate(ar) if fnmatch.fnmatch(x,pattern)]
    return ar.take(indices)

print(func(a,"??dd"))

:

['zzdd' 'zddd' 'zndd' 'nddd' 'nndd' 'dddd']

( ):

from numpy import array
import re

def func(ar,pattern):
    indices = [i for i,x in enumerate(ar) if re.match(pattern,x)]
    return ar.take(indices)

print(func(a,"..dd$"))

Answer 4

import fnmatch
import numpy as np
a = ['zzzz', 'zzzd', 'zzdd', 'zddd', 'dddn', 'ddnz', 'dnzn', 'nznz',
       'znzn', 'nznd', 'zndd', 'nddd', 'ddnn', 'dnnn', 'nnnz', 'nnzn',
       'nznn', 'znnn', 'nnnn', 'nnnd', 'nndd', 'dddz', 'ddzn', 'dznn',
       'znnz', 'nnzz', 'nzzz', 'zzzn', 'zznn', 'dddd', 'dnnd']


b=[]
for item in a:
    if fnmatch.fnmatch(item, "z*dd"):
        b.append(item)
print b

['zzdd', 'zddd', 'zndd']

Answer 5

Python .endswith(). , , . , :

i = 0
while i < len(a) :
   if a[i].endswith("dd") :
      print(a[i])
   i += 1