Code like this
def IsEven(n):
if n%2==0:
return "Is even"
else:
return "Is odd"
can be converted to a function like this
def isEven(n):
return ["Is even","Is odd"][n%2==0]
The question is if code like this:
def intervalsToOutput(n):
intervals=[(x1,x2), (x3,x4), (x5,x6), ... (xn,xn+1)]
if x1<=n<=x2:
return "in first interval"
elif x3<=n<=x4:
return "in second interval"
elif x5<=n<=x6:
return "in third interval"
...
elif xn<=n<=xn+1:
return "in last interval"
... can be replaced efficiently with a function like this:
(assuming the intervals (xi,xi+1) are not overlapping, and are sorted by xi)
def intervalsToOutput(n):
intervals=[(x1,x2), (x3,x4), (x5,x6), ... (xn,xn+1)]
answer=["in first interval","in second interval","in third interval",...,"in last interval"]
return answer[index of (n in Interval)]
The best I made (looking for speed) is
def intervalsToOutput(n):
intervals=[(x1,x2), (x3,x4), (x5,x6), ... (xn,xn+1)]
answer=["in first interval","in second interval","in third interval",...,"in last interval"]
import bisect as bisect
return answer[bisect.bisect_left(intervals, (n, )) - 1]# -1 because it is a zero based list
What bisect_left does is to find the position of n in intervals (in O(log(n)) time) by comparing (n,) with (xn,xn+1)
But the purpose of bisect is to find the insertion place, so it fails for any n which is x_{i+1}<n<=x_{j}
... (x_i, x_{i+1}), (x_j, x_{j+1}) ...
[x_i______x_i+1] fails on this interval x_j [________x_j+1]
EDIT:
This is a solution using intervaltree instead of bisect, but it has some overhead, because intervaltree returns a set, which may be empty
(I also accept faster/more elegant solutions based on this code)
#generate intervals and answers
import random
min=0
max=20
numIntervals=6
def Intervals(min,max,n):
randInts = random.sample(range(min, max), n * 2)
randInts.sort()
return [(x1,x2) for x1,x2 in zip(randInts[::2], randInts[1::2])]
intervals=Intervals(min,max,numIntervals)
answer=[f"In {n}th interval" for n in range(len(intervals))]
#Implement solution
import intervaltree
tree = intervaltree.IntervalTree()
[tree.addi(i[0],i[1],a) for i,a in zip(intervals,answer)]
def intervalsToOutput(x):
answer=tree.at(x)
if len(answer)>0:
return answer.pop().data
return "value not found"
print(intervals)
[print(f"{x} is ",intervalsToOutput(x)) for x in random.sample(range(min,max),numIntervals)]
I would do this:
intervals=[(1,2),(6,8),(22,45),(101,110)]
tgt=27
idx=next(i for i,t in enumerate(intervals) if t[0]<=tgt<=t[1])
>>> idx
2
If you want to catch a not-found condition, either use try / except:
try:
idx=next(i for i,t in enumerate(intervals) if t[0]<=tgt<=t[1])
except StopIteration:
# not found
Or use the default form of next:
idx=next((i for i,t in enumerate(intervals) if t[0]<=tgt<=t[1]), None)
Fair point regarding this method being slow. Here is a monumentally faster way.
You can take the source of bisect and modify with custom logic to inspect each tuple in the list see if the condition of t[0] <= x <= t[1] and return None if that condition cannot be satisfied.
Given a list of tuples of this form:
[(0, 3), (5, 10), (13, 17)] ... [(69735, 69739), (69742, 69746), (69749, 69752)]
You can have a custom bisect search to find a tuple that satisfies the condition stated:
def bisect_search(a, x, lo=0, hi=None):
if lo < 0:
raise ValueError('lo must be non-negative')
if hi is None or hi>len(a):
hi = len(a)
while lo < hi:
mid = (lo + hi) // 2
f_mid=-1 if x<a[mid][0] else 1 if x>a[mid][-1] else 0
if f_mid < 0:
hi = mid
elif f_mid > 0:
lo = mid + 1
else:
return mid
return None
Here is a benchmark for those two methods:
import random
import time
def gen_tuples(cnt, span=(1,5)):
li=[]
lo,hi=0,random.randint(*span)
for _ in range(cnt):
li.append((lo,hi))
lo=li[-1][-1]+random.randint(*span)
hi=lo+random.randint(span[0]+2,span[-1])
rando=random.randint(cnt//2,cnt-1)
to_find=li[rando][random.randint(0,len(li[rando])-1)]
return (rando,to_find,li)
def next_(a, x):
return next((i for i,t in enumerate(a) if t[0] <= x <=t [1]), None)
def bisect_search(a, x, lo=0, hi=None):
if lo < 0:
raise ValueError('lo must be non-negative')
if hi is None or hi>len(a):
hi = len(a)
while lo < hi:
mid = (lo + hi) // 2
f_mid=-1 if x<a[mid][0] else 1 if x>a[mid][-1] else 0
if f_mid < 0:
hi = mid
elif f_mid > 0:
lo = mid + 1
else:
return mid
return None
def cmpthese(funcs, args=(), cnt=1000, rate=True, micro=True):
"""Generate a Perl style function benchmark"""
def pprint_table(table):
"""Perl style table output"""
def format_field(field, fmt='{:,.0f}'):
if type(field) is str: return field
if type(field) is tuple: return field[1].format(field[0])
return fmt.format(field)
def get_max_col_w(table, index):
return max([len(format_field(row[index])) for row in table])
col_paddings=[get_max_col_w(table, i) for i in range(len(table[0]))]
for i,row in enumerate(table):
# left col
row_tab=[row[0].ljust(col_paddings[0])]
# rest of the cols
row_tab+=[format_field(row[j]).rjust(col_paddings[j]) for j in range(1,len(row))]
print(' '.join(row_tab))
results={}
for i in range(cnt):
for f in funcs:
start=time.perf_counter_ns()
f(*args)
stop=time.perf_counter_ns()
results.setdefault(f.__name__, []).append(stop-start)
results={k:float(sum(v))/len(v) for k,v in results.items()}
fastest=sorted(results,key=results.get, reverse=True)
table=[['']]
if rate: table[0].append('rate/sec')
if micro: table[0].append('\u03bcsec/pass')
table[0].extend(fastest)
for e in fastest:
tmp=[e]
if rate:
tmp.append('{:,}'.format(int(round(float(cnt)*1000000.0/results[e]))))
if micro:
tmp.append('{:,.1f}'.format(results[e]/float(cnt)))
for x in fastest:
if x==e: tmp.append('--')
else: tmp.append('{:.1%}'.format((results[x]-results[e])/results[e]))
table.append(tmp)
pprint_table(table)
if __name__=='__main__':
import sys
print(sys.version)
cases=(
('small, found', True, 100),
('small, not found', False, 100),
('large, found', True, 10000),
('large, not found', False, 10000)
)
for txt, f, cnt in cases:
rando,to_find,li=gen_tuples(cnt)
tgt=to_find if f else -1
args=(li, tgt)
f1,f2=bisect_search(*args), next_(*args)
print(f'\n{txt}: {f1}, {f2}')
cmpthese([next_,bisect_search], args)
Prints:
3.10.2 (main, Feb 2 2022, 06:19:27) [Clang 13.0.0 (clang-1300.0.29.3)]
small, found: 84, 84
rate/sec μsec/pass next_ bisect_search
next_ 121,533 8.2 -- -82.0%
bisect_search 675,074 1.5 455.5% --
small, not found: None, None
rate/sec μsec/pass next_ bisect_search
next_ 155,885 6.4 -- -81.6%
bisect_search 847,744 1.2 443.8% --
large, found: 8824, 8824
rate/sec μsec/pass next_ bisect_search
next_ 1,232 811.9 -- -99.6%
bisect_search 299,677 3.3 24230.9% --
large, not found: None, None
rate/sec μsec/pass next_ bisect_search
next_ 1,524 656.0 -- -99.6%
bisect_search 340,112 2.9 22211.5% --
Which shows that the bisect_search method is substantially faster.
Note: It is tempting to try and use the the key in Python 3.10 bisect to do the same thing. This is not possible since the upper end of the range of the target tuple is not known and the method of (x,) is only comparing to the bottom of the range of the target tuple.
Well here's one way, but don't do this (for readability, and probably performance). You should really just chain if,elif,andelse statements (or use a for loop):
def really_esoteric(n):
return ["first", "second", "n"][
[n in range(0,10),
n in range(11,16),
n in range(16,30)].index(True)
]
REPL:
>>> really_esoteric(5)
'first'
>>> really_esoteric(13)
'second'
>>> really_esoteric(20)
'n'
Another way of going about this:
from bisect import bisect
def search_ranges(n, intervals):
# Get the index on an insertion
index = bisect(intervals, (n,))
# Get the low and high values for the interval
lower_low, lower_high = intervals[index - 1]
upper_low, upper_high = intervals[index % len(intervals)]
# Test if n is in the interval
if n in range(lower_low, lower_high):
return f"{n} in {intervals[index-1]}"
elif n in range(upper_low, upper_high):
return f"{n} in {intervals[index]}"
else:
return f"{n} not in an interval"
print(search_ranges(0, [(0, 10)]))
print(search_ranges(4, [(0, 10)]))
print(search_ranges(11, [(0, 10), (11,20)]))
print(search_ranges(12, [(0, 10), (15,20)]))
print(search_ranges(19, [(0, 10), (11,20)]))
print(search_ranges(11, [(0, 10), (11, 20), (21, 30)]))
Outputs:
0 in (0, 10)
4 in (0, 10)
11 in (11, 20)
12 not in an interval
19 in (11, 20)
11 in (11, 20)
Are the intervals random? i.e. if the range is 0-10 and each there are 5 intervals (0-2)(2-4)(4-6)(6-8)(8-10)
Just do something like
def func(n):
spacing = 2
return n // 2
In Python or NumPy, what is the best way to find out the first occurrence of a subarray?
For example, I have
a = [1, 2, 3, 4, 5, 6]
b = [2, 3, 4]
What is the fastest way (run-time-wise) to find out where b occurs in a? I understand for strings this is extremely easy, but what about for a list or numpy ndarray?
Thanks a lot!
[EDITED] I prefer the numpy solution, since from my experience numpy vectorization is much faster than Python list comprehension. Meanwhile, the big array is huge, so I don't want to convert it into a string; that will be (too) long.
I'm assuming you're looking for a numpy-specific solution, rather than a simple list comprehension or for loop. One straightforward approach is to use the rolling window technique to search for windows of the appropriate size.
This approach is simple, works correctly, and is much faster than any pure Python solution. It should be sufficient for many use cases. However, it is not the most efficient approach possible, for a number of reasons. For an approach that is more complicated, but asymptotically optimal in the expected case, see the numba-based rolling hash implementation in norok2's answer.
Here's the rolling_window function:
>>> def rolling_window(a, size):
... shape = a.shape[:-1] + (a.shape[-1] - size + 1, size)
... strides = a.strides + (a. strides[-1],)
... return numpy.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
...
Then you could do something like
>>> a = numpy.arange(10)
>>> numpy.random.shuffle(a)
>>> a
array([7, 3, 6, 8, 4, 0, 9, 2, 1, 5])
>>> rolling_window(a, 3) == [8, 4, 0]
array([[False, False, False],
[False, False, False],
[False, False, False],
[ True, True, True],
[False, False, False],
[False, False, False],
[False, False, False],
[False, False, False]], dtype=bool)
To make this really useful, you'd have to reduce it along axis 1 using all:
>>> numpy.all(rolling_window(a, 3) == [8, 4, 0], axis=1)
array([False, False, False, True, False, False, False, False], dtype=bool)
Then you could use that however you'd use a boolean array. A simple way to get the index out:
>>> bool_indices = numpy.all(rolling_window(a, 3) == [8, 4, 0], axis=1)
>>> numpy.mgrid[0:len(bool_indices)][bool_indices]
array([3])
For lists you could adapt one of these rolling window iterators to use a similar approach.
For very large arrays and subarrays, you could save memory like this:
>>> windows = rolling_window(a, 3)
>>> sub = [8, 4, 0]
>>> hits = numpy.ones((len(a) - len(sub) + 1,), dtype=bool)
>>> for i, x in enumerate(sub):
... hits &= numpy.in1d(windows[:,i], [x])
...
>>> hits
array([False, False, False, True, False, False, False, False], dtype=bool)
>>> hits.nonzero()
(array([3]),)
On the other hand, this will probably be somewhat slower.
The following code should work:
[x for x in xrange(len(a)) if a[x:x+len(b)] == b]
Returns the index at which the pattern starts.
(EDITED to include a deeper discussion, better code and more benchmarks)
Summary
For raw speed and efficiency, one can use a Cython or Numba accelerated version (when the input is a Python sequence or a NumPy array, respectively) of one of the classical algorithms.
The recommended approaches are:
find_kmp_cy() for Python sequences (list, tuple, etc.)
find_kmp_nb() for NumPy arrays
Other efficient approaches, are find_rk_cy() and find_rk_nb() which, are more memory efficient but are not guaranteed to run in linear time.
If Cython / Numba are not available, again both find_kmp() and find_rk() are a good all-around solution for most use cases, although in the average case and for Python sequences, the naïve approach, in some form, notably find_pivot(), may be faster. For NumPy arrays, find_conv() (from #Jaime answer) outperforms any non-accelerated naïve approach.
(Full code is below, and here and there.)
Theory
This is a classical problem in computer science that goes by the name of string-searching or string matching problem.
The naive approach, based on two nested loops, has a computational complexity of O(n + m) on average, but worst case is O(n m).
Over the years, a number of alternative approaches have been developed which guarantee a better worst case performances.
Of the classical algorithms, the ones that can be best suited to generic sequences (since they do not rely on an alphabet) are:
the naïve algorithm (basically consisting of two nested loops)
the Knuth–Morris–Pratt (KMP) algorithm
the Rabin-Karp (RK) algorithm
This last algorithm relies on the computation of a rolling hash for its efficiency and therefore may require some additional knowledge of the input for optimal performance.
Eventually, it is best suited for homogeneous data, like for example numeric arrays.
A notable example of numeric arrays in Python is, of course, NumPy arrays.
Remarks
The naïve algorithm, by being so simple, lends itself to different implementations with various degrees of run-time speed in Python.
The other algorithms are less flexible in what can be optimized via language tricks.
Explicit looping in Python may be a speed bottleneck and several tricks can be used to perform the looping outside of the interpreter.
Cython is especially good at speeding up explicit loops for generic Python code.
Numba is especially good at speeding up explicit loops on NumPy arrays.
This is an excellent use-case for generators, so all the code will be using those instead of regular functions.
Python Sequences (list, tuple, etc.)
Based on the Naïve Algorithm
find_loop(), find_loop_cy() and find_loop_nb() which are the explicit-loop only implementation in pure Python, Cython and with Numba JITing respectively. Note the forceobj=True in the Numba version, which is required because we are using Python object inputs.
def find_loop(seq, subseq):
n = len(seq)
m = len(subseq)
for i in range(n - m + 1):
found = True
for j in range(m):
if seq[i + j] != subseq[j]:
found = False
break
if found:
yield i
%%cython -c-O3 -c-march=native -a
#cython: language_level=3, boundscheck=False, wraparound=False, initializedcheck=False, cdivision=True, infer_types=True
def find_loop_cy(seq, subseq):
cdef Py_ssize_t n = len(seq)
cdef Py_ssize_t m = len(subseq)
for i in range(n - m + 1):
found = True
for j in range(m):
if seq[i + j] != subseq[j]:
found = False
break
if found:
yield i
find_loop_nb = nb.jit(find_loop, forceobj=True)
find_loop_nb.__name__ = 'find_loop_nb'
find_all() replaces the inner loop with all() on a comprehension generator
def find_all(seq, subseq):
n = len(seq)
m = len(subseq)
for i in range(n - m + 1):
if all(seq[i + j] == subseq[j] for j in range(m)):
yield i
find_slice() replaces the inner loop with direct comparison == after slicing []
def find_slice(seq, subseq):
n = len(seq)
m = len(subseq)
for i in range(n - m + 1):
if seq[i:i + m] == subseq:
yield i
find_mix() and find_mix2() replaces the inner loop with direct comparison == after slicing [] but includes one or two additional short-circuiting on the first (and last) character which may be faster because slicing with an int is much faster than slicing with a slice().
def find_mix(seq, subseq):
n = len(seq)
m = len(subseq)
for i in range(n - m + 1):
if seq[i] == subseq[0] and seq[i:i + m] == subseq:
yield i
def find_mix2(seq, subseq):
n = len(seq)
m = len(subseq)
for i in range(n - m + 1):
if seq[i] == subseq[0] and seq[i + m - 1] == subseq[m - 1] \
and seq[i:i + m] == subseq:
yield i
find_pivot() and find_pivot2() replace the outer loop with multiple .index() call using the first item of the sub-sequence, while using slicing for the inner loop, eventually with additional short-circuiting on the last item (the first matches by construction). The multiple .index() calls are wrapped in a index_all() generator (which may be useful on its own).
def index_all(seq, item, start=0, stop=-1):
try:
n = len(seq)
if n > 0:
start %= n
stop %= n
i = start
while True:
i = seq.index(item, i)
if i <= stop:
yield i
i += 1
else:
return
else:
return
except ValueError:
pass
def find_pivot(seq, subseq):
n = len(seq)
m = len(subseq)
if m > n:
return
for i in index_all(seq, subseq[0], 0, n - m):
if seq[i:i + m] == subseq:
yield i
def find_pivot2(seq, subseq):
n = len(seq)
m = len(subseq)
if m > n:
return
for i in index_all(seq, subseq[0], 0, n - m):
if seq[i + m - 1] == subseq[m - 1] and seq[i:i + m] == subseq:
yield i
Based on Knuth–Morris–Pratt (KMP) Algorithm
find_kmp() is a plain Python implementation of the algorithm. Since there is no simple looping or places where one could use slicing with a slice(), there is not much to be done for optimization, except using Cython (Numba would require again forceobj=True which would lead to slow code).
def find_kmp(seq, subseq):
n = len(seq)
m = len(subseq)
# : compute offsets
offsets = [0] * m
j = 1
k = 0
while j < m:
if subseq[j] == subseq[k]:
k += 1
offsets[j] = k
j += 1
else:
if k != 0:
k = offsets[k - 1]
else:
offsets[j] = 0
j += 1
# : find matches
i = j = 0
while i < n:
if seq[i] == subseq[j]:
i += 1
j += 1
if j == m:
yield i - j
j = offsets[j - 1]
elif i < n and seq[i] != subseq[j]:
if j != 0:
j = offsets[j - 1]
else:
i += 1
find_kmp_cy() is Cython implementation of the algorithm where the indices use C int data type, which result in much faster code.
%%cython -c-O3 -c-march=native -a
#cython: language_level=3, boundscheck=False, wraparound=False, initializedcheck=False, cdivision=True, infer_types=True
def find_kmp_cy(seq, subseq):
cdef Py_ssize_t n = len(seq)
cdef Py_ssize_t m = len(subseq)
# : compute offsets
offsets = [0] * m
cdef Py_ssize_t j = 1
cdef Py_ssize_t k = 0
while j < m:
if subseq[j] == subseq[k]:
k += 1
offsets[j] = k
j += 1
else:
if k != 0:
k = offsets[k - 1]
else:
offsets[j] = 0
j += 1
# : find matches
cdef Py_ssize_t i = 0
j = 0
while i < n:
if seq[i] == subseq[j]:
i += 1
j += 1
if j == m:
yield i - j
j = offsets[j - 1]
elif i < n and seq[i] != subseq[j]:
if j != 0:
j = offsets[j - 1]
else:
i += 1
Based on Rabin-Karp (RK) Algorithm
find_rk() is a pure Python implementation, which relies on Python's hash() for the computation (and comparison) of the hash. Such hash is made rolling by mean of a simple sum(). The roll-over is then computed from the previous hash by subtracting the result of hash() on the just visited item seq[i - 1] and adding up the result of hash() on the newly considered item seq[i + m - 1].
def find_rk(seq, subseq):
n = len(seq)
m = len(subseq)
if seq[:m] == subseq:
yield 0
hash_subseq = sum(hash(x) for x in subseq) # compute hash
curr_hash = sum(hash(x) for x in seq[:m]) # compute hash
for i in range(1, n - m + 1):
curr_hash += hash(seq[i + m - 1]) - hash(seq[i - 1]) # update hash
if hash_subseq == curr_hash and seq[i:i + m] == subseq:
yield i
find_rk_cy() is Cython implementation of the algorithm where the indices use the appropriate C data type, which results in much faster code. Note that hash() truncates "the return value based on the bit width of the host machine."
%%cython -c-O3 -c-march=native -a
#cython: language_level=3, boundscheck=False, wraparound=False, initializedcheck=False, cdivision=True, infer_types=True
def find_rk_cy(seq, subseq):
cdef Py_ssize_t n = len(seq)
cdef Py_ssize_t m = len(subseq)
if seq[:m] == subseq:
yield 0
cdef Py_ssize_t hash_subseq = sum(hash(x) for x in subseq) # compute hash
cdef Py_ssize_t curr_hash = sum(hash(x) for x in seq[:m]) # compute hash
cdef Py_ssize_t old_item, new_item
for i in range(1, n - m + 1):
old_item = hash(seq[i - 1])
new_item = hash(seq[i + m - 1])
curr_hash += new_item - old_item # update hash
if hash_subseq == curr_hash and seq[i:i + m] == subseq:
yield i
Benchmarks
The above functions are evaluated on two inputs:
random inputs
def gen_input(n, k=2):
return tuple(random.randint(0, k - 1) for _ in range(n))
(almost) worst inputs for the naïve algorithm
def gen_input_worst(n, k=-2):
result = [0] * n
result[k] = 1
return tuple(result)
The subseq has fixed size (32).
Since there are so many alternatives, two separate grouping have been done and some solutions with very small variations and almost identical timings have been omitted (i.e. find_mix2() and find_pivot2()).
For each group both inputs are tested.
For each benchmark the full plot and a zoom on the fastest approach is provided.
Naïve on Random
Naïve on Worst
Other on Random
Other on Worst
(Full code is available here.)
NumPy Arrays
Based on the Naïve Algorithm
find_loop(), find_loop_cy() and find_loop_nb() which are the explicit-loop only implementation in pure Python, Cython and with Numba JITing respectively. The code for the first two are the same as above and hence omitted. find_loop_nb() now enjoys fast JIT compilation. The inner loop has been written in a separate function because it can then be reused for find_rk_nb() (calling Numba functions inside Numba functions does not incur in the function call penalty typical of Python).
#nb.jit
def _is_equal_nb(seq, subseq, m, i):
for j in range(m):
if seq[i + j] != subseq[j]:
return False
return True
#nb.jit
def find_loop_nb(seq, subseq):
n = len(seq)
m = len(subseq)
for i in range(n - m + 1):
if _is_equal_nb(seq, subseq, m, i):
yield i
find_all() is the same as above, while find_slice(), find_mix() and find_mix2() are almost identical to the above, the only difference is that seq[i:i + m] == subseq is now the argument of np.all(): np.all(seq[i:i + m] == subseq).
find_pivot() and find_pivot2() share the same ideas as above, except that now uses np.where() instead of index_all() and the need for enclosing the array equality inside an np.all() call.
def find_pivot(seq, subseq):
n = len(seq)
m = len(subseq)
if m > n:
return
max_i = n - m
for i in np.where(seq == subseq[0])[0]:
if i > max_i:
return
elif np.all(seq[i:i + m] == subseq):
yield i
def find_pivot2(seq, subseq):
n = len(seq)
m = len(subseq)
if m > n:
return
max_i = n - m
for i in np.where(seq == subseq[0])[0]:
if i > max_i:
return
elif seq[i + m - 1] == subseq[m - 1] \
and np.all(seq[i:i + m] == subseq):
yield i
find_rolling() express the looping via a rolling window and the matching is checked with np.all(). This vectorizes all the looping at the expenses of creating large temporary objects, while still substantially appling the naïve algorithm. (The approach is from #senderle answer).
def rolling_window(arr, size):
shape = arr.shape[:-1] + (arr.shape[-1] - size + 1, size)
strides = arr.strides + (arr.strides[-1],)
return np.lib.stride_tricks.as_strided(arr, shape=shape, strides=strides)
def find_rolling(seq, subseq):
bool_indices = np.all(rolling_window(seq, len(subseq)) == subseq, axis=1)
yield from np.mgrid[0:len(bool_indices)][bool_indices]
find_rolling2() is a slightly more memory efficient variation of the above, where the vectorization is only partial and one explicit looping (along the expected shortest dimension -- the length of subseq) is kept. (The approach is also from #senderle answer).
def find_rolling2(seq, subseq):
windows = rolling_window(seq, len(subseq))
hits = np.ones((len(seq) - len(subseq) + 1,), dtype=bool)
for i, x in enumerate(subseq):
hits &= np.in1d(windows[:, i], [x])
yield from hits.nonzero()[0]
Based on Knuth–Morris–Pratt (KMP) Algorithm
find_kmp() is the same as above, while find_kmp_nb() is a straightforward JIT-compilation of that.
find_kmp_nb = nb.jit(find_kmp)
find_kmp_nb.__name__ = 'find_kmp_nb'
Based on Rabin-Karp (RK) Algorithm
find_rk() is the same as the above, except that again seq[i:i + m] == subseq is enclosed in an np.all() call.
find_rk_nb() is the Numba accelerated version of the above. Uses _is_equal_nb() defined earlier to definitively determine a match, while for the hashing, it uses a Numba accelerated sum_hash_nb() function whose definition is pretty straightforward.
#nb.jit
def sum_hash_nb(arr):
result = 0
for x in arr:
result += hash(x)
return result
#nb.jit
def find_rk_nb(seq, subseq):
n = len(seq)
m = len(subseq)
if _is_equal_nb(seq, subseq, m, 0):
yield 0
hash_subseq = sum_hash_nb(subseq) # compute hash
curr_hash = sum_hash_nb(seq[:m]) # compute hash
for i in range(1, n - m + 1):
curr_hash += hash(seq[i + m - 1]) - hash(seq[i - 1]) # update hash
if hash_subseq == curr_hash and _is_equal_nb(seq, subseq, m, i):
yield i
find_conv() uses a pseudo Rabin-Karp method, where initial candidates are hashed using the np.dot() product and located on the convolution between seq and subseq with np.where(). The approach is pseudo because, while it still uses hashing to identify probable candidates, it is may not be regarded as a rolling hash (it depends on the actual implementation of np.correlate()). Also, it needs to create a temporary array the size of the input. (The approach is from #Jaime answer).
def find_conv(seq, subseq):
target = np.dot(subseq, subseq)
candidates = np.where(np.correlate(seq, subseq, mode='valid') == target)[0]
check = candidates[:, np.newaxis] + np.arange(len(subseq))
mask = np.all((np.take(seq, check) == subseq), axis=-1)
yield from candidates[mask]
Benchmarks
Like before, the above functions are evaluated on two inputs:
random inputs
def gen_input(n, k=2):
return np.random.randint(0, k, n)
(almost) worst inputs for the naïve algorithm
def gen_input_worst(n, k=-2):
result = np.zeros(n, dtype=int)
result[k] = 1
return result
The subseq has fixed size (32).
This plots follow the same scheme as before, summarized below for convenience.
Since there are so many alternatives, two separate grouping have been done and some solutions with very small variations and almost identical timings have been omitted (i.e. find_mix2() and find_pivot2()).
For each group both inputs are tested.
For each benchmark the full plot and a zoom on the fastest approach is provided.
Naïve on Random
Naïve on Worst
Other on Random
Other on Worst
(Full code is available here.)
A convolution based approach, that should be more memory efficient than the stride_tricks based approach:
def find_subsequence(seq, subseq):
target = np.dot(subseq, subseq)
candidates = np.where(np.correlate(seq,
subseq, mode='valid') == target)[0]
# some of the candidates entries may be false positives, double check
check = candidates[:, np.newaxis] + np.arange(len(subseq))
mask = np.all((np.take(seq, check) == subseq), axis=-1)
return candidates[mask]
With really big arrays it may not be possible to use a stride_tricks approach, but this one still works:
haystack = np.random.randint(1000, size=(1e6))
needle = np.random.randint(1000, size=(100,))
# Hide 10 needles in the haystack
place = np.random.randint(1e6 - 100 + 1, size=10)
for idx in place:
haystack[idx:idx+100] = needle
In [3]: find_subsequence(haystack, needle)
Out[3]:
array([253824, 321497, 414169, 456777, 635055, 879149, 884282, 954848,
961100, 973481], dtype=int64)
In [4]: np.all(np.sort(place) == find_subsequence(haystack, needle))
Out[4]: True
In [5]: %timeit find_subsequence(haystack, needle)
10 loops, best of 3: 79.2 ms per loop
you can call tostring() method to convert an array to string, and then you can use fast string search. this method maybe faster when you have many subarray to check.
import numpy as np
a = np.array([1,2,3,4,5,6])
b = np.array([2,3,4])
print a.tostring().index(b.tostring())//a.itemsize
Another try, but I'm sure there is more pythonic & efficent way to do that ...
def array_match(a, b):
for i in xrange(0, len(a)-len(b)+1):
if a[i:i+len(b)] == b:
return i
return None
a = [1, 2, 3, 4, 5, 6]
b = [2, 3, 4]
print array_match(a,b)
1
(This first answer was not in scope of the question, as cdhowie mentionned)
set(a) & set(b) == set(b)
Here is a rather straight-forward option:
def first_subarray(full_array, sub_array):
n = len(full_array)
k = len(sub_array)
matches = np.argwhere([np.all(full_array[start_ix:start_ix+k] == sub_array)
for start_ix in range(0, n-k+1)])
return matches[0]
Then using the original a, b vectors we get:
a = [1, 2, 3, 4, 5, 6]
b = [2, 3, 4]
first_subarray(a, b)
Out[44]:
array([1], dtype=int64)
Quick comparison of three of the proposed solutions (average time of 100 iteration for randomly created vectors.):
import time
import collections
import numpy as np
def function_1(seq, sub):
# direct comparison
seq = list(seq)
sub = list(sub)
return [i for i in range(len(seq) - len(sub)) if seq[i:i+len(sub)] == sub]
def function_2(seq, sub):
# Jamie's solution
target = np.dot(sub, sub)
candidates = np.where(np.correlate(seq, sub, mode='valid') == target)[0]
check = candidates[:, np.newaxis] + np.arange(len(sub))
mask = np.all((np.take(seq, check) == sub), axis=-1)
return candidates[mask]
def function_3(seq, sub):
# HYRY solution
return seq.tostring().index(sub.tostring())//seq.itemsize
# --- assessment time performance
N = 100
seq = np.random.choice([0, 1, 2, 3, 4, 5, 6], 3000)
sub = np.array([1, 2, 3])
tim = collections.OrderedDict()
tim.update({function_1: 0.})
tim.update({function_2: 0.})
tim.update({function_3: 0.})
for function in tim.keys():
for _ in range(N):
seq = np.random.choice([0, 1, 2, 3, 4], 3000)
sub = np.array([1, 2, 3])
start = time.time()
function(seq, sub)
end = time.time()
tim[function] += end - start
timer_dict = collections.OrderedDict()
for key, val in tim.items():
timer_dict.update({key.__name__: val / N})
print(timer_dict)
Which would result (on my old machine) in:
OrderedDict([
('function_1', 0.0008518099784851074),
('function_2', 8.157730102539063e-05),
('function_3', 6.124973297119141e-06)
])
First, convert the list to string.
a = ''.join(str(i) for i in a)
b = ''.join(str(i) for i in b)
After converting to string, you can easily find the index of substring with the following string function.
a.index(b)
Cheers!!
I am trying to solving the "Counting Change" problem with memorization.
Consider the following problem: How many different ways can we make change of $1.00, given half-dollars, quarters, dimes, nickels, and pennies? More generally, can we write a function to compute the number of ways to change any given amount of money using any set of currency denominations?
And the intuitive solution with recursoin.
The number of ways to change an amount a using n kinds of coins equals
the number of ways to change a using all but the first kind of coin, plus
the number of ways to change the smaller amount a - d using all n kinds of coins, where d is the denomination of the first kind of coin.
#+BEGIN_SRC python :results output
# cache = {} # add cache
def count_change(a, kinds=(50, 25, 10, 5, 1)):
"""Return the number of ways to change amount a using coin kinds."""
if a == 0:
return 1
if a < 0 or len(kinds) == 0:
return 0
d = kinds[0] # d for digit
return count_change(a, kinds[1:]) + count_change(a - d, kinds)
print(count_change(100))
#+END_SRC
#+RESULTS:
: 292
I try to take advantage of memorization,
Signature: count_change(a, kinds=(50, 25, 10, 5, 1))
Source:
def count_change(a, kinds=(50, 25, 10, 5, 1)):
"""Return the number of ways to change amount a using coin kinds."""
if a == 0:
return 1
if a < 0 or len(kinds) == 0:
return 0
d = kinds[0]
cache[a] = count_change(a, kinds[1:]) + count_change(a - d, kinds)
return cache[a]
It works properly for small number like
In [17]: count_change(120)
Out[17]: 494
work on big numbers
In [18]: count_change(11000)
---------------------------------------------------------------------------
RecursionError Traceback (most recent call last)
<ipython-input-18-52ba30c71509> in <module>
----> 1 count_change(11000)
/tmp/ipython_edit_h0rppahk/ipython_edit_uxh2u429.py in count_change(a, kinds)
9 return 0
10 d = kinds[0]
---> 11 cache[a] = count_change(a, kinds[1:]) + count_change(a - d, kinds)
12 return cache[a]
... last 1 frames repeated, from the frame below ...
/tmp/ipython_edit_h0rppahk/ipython_edit_uxh2u429.py in count_change(a, kinds)
9 return 0
10 d = kinds[0]
---> 11 cache[a] = count_change(a, kinds[1:]) + count_change(a - d, kinds)
12 return cache[a]
RecursionError: maximum recursion depth exceeded in comparison
What's the problem with memorization solution?
In the memoized version, the count_change function has to take into account the highest index of coin you can use when you make the recursive call, so that you can use the already calculated values ...
def count_change(n, k, kinds):
if n < 0:
return 0
if (n, k) in cache:
return cache[n,k]
if k == 0:
v = 1
else:
v = count_change(n-kinds[k], k, kinds) + count_change(n, k-1, kinds)
cache[n,k] = v
return v
You can try :
cache = {}
count_change(120,4, [1, 5, 10, 25, 50])
gives 494
while :
cache = {}
count_change(11000,4, [1, 5, 10, 25, 50])
outputs: 9930221951