How can I slice each element of a numpy array of strings?

Question

Numpy has some very useful string operations, which vectorize the usual Python string operations.

Compared to these operation and to pandas.str, the num

Answer 1

Here's a vectorized approach -

def slicer_vectorized(a,start,end):
    b = a.view((str,1)).reshape(len(a),-1)[:,start:end]
    return np.fromstring(b.tostring(),dtype=(str,end-start))

Sample run -

In [68]: a = np.array(['hello', 'how', 'are', 'you'])

In [69]: slicer_vectorized(a,1,3)
Out[69]: 
array(['el', 'ow', 're', 'ou'], 
      dtype='|S2')

In [70]: slicer_vectorized(a,0,3)
Out[70]: 
array(['hel', 'how', 'are', 'you'], 
      dtype='|S3')

Runtime test -

Testing out all the approaches posted by other authors that I could run at my end and also including the vectorized approach from earlier in this post.

Here's the timings -

In [53]: # Setup input array
    ...: a = np.array(['hello', 'how', 'are', 'you'])
    ...: a = np.repeat(a,10000)
    ...: 

# @Alberto Garcia-Raboso's answer
In [54]: %timeit slicer(1, 3)(a)
10 loops, best of 3: 23.5 ms per loop

# @hapaulj's answer
In [55]: %timeit np.frompyfunc(lambda x:x[1:3],1,1)(a)
100 loops, best of 3: 11.6 ms per loop

# Using loop-comprehension
In [56]: %timeit np.array([i[1:3] for i in a])
100 loops, best of 3: 12.1 ms per loop

# From this post
In [57]: %timeit slicer_vectorized(a,1,3)
1000 loops, best of 3: 787 µs per loop

Answer 2

Most, if not all the functions in np.char apply existing str methods to each element of the array. It's a little faster than direct iteration (or vectorize) but not drastically so.

There isn't a string slicer; at least not by that sort of name. Closest is indexing with a slice:

In [274]: 'astring'[1:3]
Out[274]: 'st'
In [275]: 'astring'.__getitem__
Out[275]: <method-wrapper '__getitem__' of str object at 0xb3866c20>
In [276]: 'astring'.__getitem__(slice(1,4))
Out[276]: 'str'

An iterative approach can be with frompyfunc (which is also used by vectorize):

In [277]: a = numpy.array(['hello', 'how', 'are', 'you'])
In [278]: np.frompyfunc(lambda x:x[1:3],1,1)(a)
Out[278]: array(['el', 'ow', 're', 'ou'], dtype=object)
In [279]: np.frompyfunc(lambda x:x[1:3],1,1)(a).astype('U2')
Out[279]: 
array(['el', 'ow', 're', 'ou'], 
      dtype='<U2')

I could view it as a single character array, and slice that

In [289]: a.view('U1').reshape(4,-1)[:,1:3]
Out[289]: 
array([['e', 'l'],
       ['o', 'w'],
       ['r', 'e'],
       ['o', 'u']], 
      dtype='<U1')

I still need to figure out how to convert it back to 'U2'.

In [290]: a.view('U1').reshape(4,-1)[:,1:3].copy().view('U2')
Out[290]: 
array([['el'],
       ['ow'],
       ['re'],
       ['ou']], 
      dtype='<U2')

The initial view step shows the databuffer as Py3 characters (these would be bytes in a S or Py2 string case):

In [284]: a.view('U1')
Out[284]: 
array(['h', 'e', 'l', 'l', 'o', 'h', 'o', 'w', '', '', 'a', 'r', 'e', '',
       '', 'y', 'o', 'u', '', ''], 
      dtype='<U1')

Picking the 1:3 columns amounts to picking a.view('U1')[[1,2,6,7,11,12,16,17]] and then reshaping and view. Without getting into details, I'm not surprised that it requires a copy.

Answer 3

To solve this, so far I've been transforming the numpy array to a pandas Series and back. It is not a pretty solution, but it works and it works relatively fast.

a = numpy.array(['hello', 'how', 'are', 'you'])
pandas.Series(a).str[1:3].values
array(['el', 'ow', 're', 'ou'], dtype=object)

Answer 4

Interesting omission... I guess you can always write your own:

import numpy as np

def slicer(start=None, stop=None, step=1):
    return np.vectorize(lambda x: x[start:stop:step], otypes=[str])

a = np.array(['hello', 'how', 'are', 'you'])
print(slicer(1, 3)(a))    # => ['el' 'ow' 're' 'ou']

EDIT: Here are some benchmarks using the text of Ulysses by James Joyce. It seems the clear winner is @hpaulj's last strategy. @Divakar gets into the race improving on @hpaulj's last strategy.

import numpy as np
import requests

ulysses = requests.get('http://www.gutenberg.org/files/4300/4300-0.txt').text
a = np.array(ulysses.split())

# Ufunc
def slicer(start=None, stop=None, step=1):
    return np.vectorize(lambda x: x[start:stop:step], otypes=[str])

%timeit slicer(1, 3)(a)
# => 1 loop, best of 3: 221 ms per loop

# Non-mutating loop
def loop1(a):
    out = np.empty(len(a), dtype=object)
    for i, word in enumerate(a):
        out[i] = word[1:3]

%timeit loop1(a)
# => 1 loop, best of 3: 262 ms per loop

# Mutating loop
def loop2(a):
    for i in range(len(a)):
        a[i] = a[i][1:3]

b = a.copy()
%timeit -n 1 -r 1 loop2(b)
# 1 loop, best of 1: 285 ms per loop

# From @hpaulj's answer
%timeit np.frompyfunc(lambda x:x[1:3],1,1)(a)
# => 10 loops, best of 3: 141 ms per loop

%timeit np.frompyfunc(lambda x:x[1:3],1,1)(a).astype('U2')
# => 1 loop, best of 3: 170 ms per loop

%timeit a.view('U1').reshape(len(a),-1)[:,1:3].astype(object).sum(axis=1)
# => 10 loops, best of 3: 60.7 ms per loop

def slicer_vectorized(a,start,end):
    b = a.view('S1').reshape(len(a),-1)[:,start:end]
    return np.fromstring(b.tostring(),dtype='S'+str(end-start))

%timeit slicer_vectorized(a,1,3)
# => The slowest run took 5.34 times longer than the fastest.
#    This could mean that an intermediate result is being cached.
#    10 loops, best of 3: 16.8 ms per loop