Just doing a review of my Python class and noticed that I forgot how to do this.
def outsideIn2(lst):
'''(list)->list
Returns a new list where the middle two elements have been
removed and placed at the beginning of the result. Assume all lists are an even
length
>>> outsideIn2(['C','a','r','t','o','n'])
['r','t','C','a','o','n'] # rt moves to front
>>> outsideIn2(['H','i'])
['H','i'] # Hi moves to front so output remains the same.
>>> outsideIn2(['B','a','r','b','a','r','a',' ','A','n','n','e'])
['r','a','B','a','r','b,','a',' ','A','n','n','e'] # ra moves to front.
'''
length = len(lst)
middle1 = lst.pop((len(lst) / 2) - 1)
middle2 = lst.pop((len(lst) / 2) + 1)
lst.insert([0], middle1)
lst.insert([1], middle2)
return lst
I'm getting this error:
middle1 = lst.pop((len(lst) / 2) - 1)
TypeError: integer argument expected, got float
What am I doing wrong?
When you upgraded to Python 3, the '/' operator changed from giving you integer division to real division. Switch to '//' operator.
You can use // operator:
middle1 = lst.pop((len(lst) // 2) - 1)
The other answers explained why you are getting the error. You need to use // instead of / (also, just for the record, you need to give list.insert integers, not lists).
However, I'd like to suggest a different approach that uses Explain Python's slice notation:
def outsideIn2(lst):
x = len(lst)//2
return lst[x-1:x+1]+lst[:x-1]+lst[x+1:]
This method should be significantly faster than usinglist.pop and list.insert.
As proof, I made the below script to compare the two methods with timeit.timeit:
from timeit import timeit
def outsideIn2(lst):
length = len(lst)
middle1 = lst.pop((len(lst) // 2) - 1)
middle2 = lst.pop((len(lst) // 2) + 1)
lst.insert(0, middle1)
lst.insert(1, middle2)
return lst
print(timeit("outsideIn2(['B','a','r','b','a','r','a',' ','A','n','n','e'])", "from __main__ import outsideIn2"))
def outsideIn2(lst):
x = len(lst)//2
return lst[x-1:x+1]+lst[:x-1]+lst[x+1:]
print(timeit("outsideIn2(['B','a','r','b','a','r','a',' ','A','n','n','e'])", "from __main__ import outsideIn2"))
The results were as follows:
6.255111473664949
4.465956427423038
As you can see, my proposed method was ~2 seconds faster. However, you can run more tests if you would like to validate mine.
Using pop and insert (especially inserting at positions 0 and 1) can be fairly slow with Python lists. Since the underlying storage for the list is an array, inserting at position 0 means that the element at position n-1 has to be moved to position n, then the element at n-2 has to be moved to n-1 and so on. pop has to do the same in reverse. So imagine in your little method how many element moves must be done. Roughly:
pop #1 - move n/2 elements
pop #2 - move n/2 elements
insert 0 - move n elements
insert 1 - move n elements
So approximately 3n moves are done in this code.
Breaking the list into 3 slices and reassembling a new list may be more optimal:
def outsideIn2(lst):
midstart = len(lst)//2-1
left,mid,right = lst[0:midstart], lst[midstart:midstart+2], lst[midstart+2:]
return mid+left+right
Plus you won't run into any weird issues by pop changing the length of the list between the first and second call to pop. And the slices implicitly guard against index errors when you get a list that is shorter than 2 characters.
Related
I could do this in brute force, but I was hoping there was clever coding, or perhaps an existing function, or something I am not realising...
So some examples of numbers I want:
00000000001111110000
11111100000000000000
01010101010100000000
10101010101000000000
00100100100100100100
The full permutation. Except with results that have ONLY six 1's. Not more. Not less. 64 or 32 bits would be ideal. 16 bits if that provides an answer.
I think what you need here is using the itertools module.
BAD SOLUTION
But you need to be careful, for instance, using something like permutations would just work for very small inputs. ie:
Something like the below would give you a binary representation:
>>> ["".join(v) for v in set(itertools.permutations(["1"]*2+["0"]*3))]
['11000', '01001', '00101', '00011', '10010', '01100', '01010', '10001', '00110', '10100']
then just getting decimal representation of those number:
>>> [int("".join(v), 16) for v in set(itertools.permutations(["1"]*2+["0"]*3))]
[69632, 4097, 257, 17, 65552, 4352, 4112, 65537, 272, 65792]
if you wanted 32bits with 6 ones and 26 zeroes, you'd use:
>>> [int("".join(v), 16) for v in set(itertools.permutations(["1"]*6+["0"]*26))]
but this computation would take a supercomputer to deal with (32! = 263130836933693530167218012160000000 )
DECENT SOLUTION
So a more clever way to do it is using combinations, maybe something like this:
import itertools
num_bits = 32
num_ones = 6
lst = [
f"{sum([2**vv for vv in v]):b}".zfill(num_bits)
for v in list(itertools.combinations(range(num_bits), num_ones))
]
print(len(lst))
this would tell us there is 906192 numbers with 6 ones in the whole spectrum of 32bits numbers.
CREDITS:
Credits for this answer go to #Mark Dickinson who pointed out using permutations was unfeasible and suggested the usage of combinations
Well I am not a Python coder so I can not post a valid code for you. Instead I can do a C++ one...
If you look at your problem you set 6 bits and many zeros ... so I would approach this by 6 nested for loops computing all the possible 1s position and set the bits...
Something like:
for (i0= 0;i0<32-5;i0++)
for (i1=i0+1;i1<32-4;i1++)
for (i2=i1+1;i2<32-3;i2++)
for (i3=i2+1;i3<32-2;i3++)
for (i4=i3+1;i4<32-1;i4++)
for (i5=i4+1;i5<32-0;i5++)
// here i0,...,i5 marks the set bits positions
So the O(2^32) become to less than `~O(26.25.24.23.22.21/16) and you can not go faster than that as that would mean you miss valid solutions...
I assume you want to print the number so for speed up you can compute the number as a binary number string from the start to avoid slow conversion between string and number...
The nested for loops can be encoded as increment operation of an array (similar to bignum arithmetics)
When I put all together I got this C++ code:
int generate()
{
const int n1=6; // number of set bits
const int n=32; // number of bits
char x[n+2]; // output number string
int i[n1],j,cnt; // nested for loops iterator variables and found solutions count
for (j=0;j<n;j++) x[j]='0'; x[j]='b'; j++; x[j]=0; // x = 0
for (j=0;j<n1;j++){ i[j]=j; x[i[j]]='1'; } // first solution
for (cnt=0;;)
{
// Form1->mm_log->Lines->Add(x); // here x is the valid answer to print
cnt++;
for (j=n1-1;j>=0;j--) // this emulates n1 nested for loops
{
x[i[j]]='0'; i[j]++;
if (i[j]<n-n1+j+1){ x[i[j]]='1'; break; }
}
if (j<0) break;
for (j++;j<n1;j++){ i[j]=i[j-1]+1; x[i[j]]='1'; }
}
return cnt; // found valid answers
};
When I use this with n1=6,n=32 I got this output (without printing the numbers):
cnt = 906192
and it was finished in 4.246 ms on AMD A8-5500 3.2GHz (win7 x64 32bit app no threads) which is fast enough for me...
Beware once you start outputing the numbers somewhere the speed will drop drastically. Especially if you output to console or what ever ... it might be better to buffer the output somehow like outputting 1024 string numbers at once etc... But as I mentioned before I am no Python coder so it might be already handled by the environment...
On top of all this once you will play with variable n1,n you can do the same for zeros instead of ones and use faster approach (if there is less zeros then ones use nested for loops to mark zeros instead of ones)
If the wanted solution numbers are wanted as a number (not a string) then its possible to rewrite this so the i[] or i0,..i5 holds the bitmask instead of bit positions ... instead of inc/dec you just shift left/right ... and no need for x array anymore as the number would be x = i0|...|i5 ...
You could create a counter array for positions of 1s in the number and assemble it by shifting the bits in their respective positions. I created an example below. It runs pretty fast (less than a second for 32 bits on my laptop):
bitCount = 32
oneCount = 6
maxBit = 1<<(bitCount-1)
ones = [1<<b for b in reversed(range(oneCount)) ] # start with bits on low end
ones[0] >>= 1 # shift back 1st one because it will be incremented at start of loop
index = 0
result = []
while index < len(ones):
ones[index] <<= 1 # shift one at current position
if index == 0:
number = sum(ones) # build output number
result.append(number)
if ones[index] == maxBit:
index += 1 # go to next position when bit reaches max
elif index > 0:
index -= 1 # return to previous position
ones[index] = ones[index+1] # and prepare it to move up (relative to next)
64 bits takes about a minute, roughly proportional to the number of values that are output. O(n)
The same approach can be expressed more concisely in a recursive generator function which will allow more efficient use of the bit patterns:
def genOneBits(bitcount=32,onecount=6):
for bitPos in range(onecount-1,bitcount):
value = 1<<bitPos
if onecount == 1: yield value; continue
for otherBits in genOneBits(bitPos,onecount-1):
yield value + otherBits
result = [ n for n in genOneBits(32,6) ]
This is not faster when you get all the numbers but it allows partial access to the list without going through all values.
If you need direct access to the Nth bit pattern (e.g. to get a random one-bits pattern), you can use the following function. It works like indexing a list but without having to generate the list of patterns.
def numOneBits(bitcount=32,onecount=6):
def factorial(X): return 1 if X < 2 else X * factorial(X-1)
return factorial(bitcount)//factorial(onecount)//factorial(bitcount-onecount)
def nthOneBits(N,bitcount=32,onecount=6):
if onecount == 1: return 1<<N
bitPos = 0
while bitPos<=bitcount-onecount:
group = numOneBits(bitcount-bitPos-1,onecount-1)
if N < group: break
N -= group
bitPos += 1
if bitPos>bitcount-onecount: return None
result = 1<<bitPos
result |= nthOneBits(N,bitcount-bitPos-1,onecount-1)<<(bitPos+1)
return result
# bit pattern at position 1000:
nthOneBit(1000) # --> 10485799 (00000000101000000000000000100111)
This allows you to get the bit patterns on very large integers that would be impossible to generate completely:
nthOneBits(10000, bitcount=256, onecount=9)
# 77371252457588066994880639
# 100000000000000000000000000000000001000000000000000000000000000000000000000000001111111
It is worth noting that the pattern order does not follow the numerical order of the corresponding numbers
Although nthOneBits() can produce any pattern instantly, it is much slower than the other functions when mass producing patterns. If you need to manipulate them sequentially, you should go for the generator function instead of looping on nthOneBits().
Also, it should be fairly easy to tweak the generator to have it start at a specific pattern so you could get the best of both approaches.
Finally, it may be useful to obtain then next bit pattern given a known pattern. This is what the following function does:
def nextOneBits(N=0,bitcount=32,onecount=6):
if N == 0: return (1<<onecount)-1
bitPositions = []
for pos in range(bitcount):
bit = N%2
N //= 2
if bit==1: bitPositions.insert(0,pos)
index = 0
result = None
while index < onecount:
bitPositions[index] += 1
if bitPositions[index] == bitcount:
index += 1
continue
if index == 0:
result = sum( 1<<bp for bp in bitPositions )
break
if index > 0:
index -= 1
bitPositions[index] = bitPositions[index+1]
return result
nthOneBits(12) #--> 131103 00000000000000100000000000011111
nextOneBits(131103) #--> 262175 00000000000001000000000000011111 5.7ns
nthOneBits(13) #--> 262175 00000000000001000000000000011111 49.2ns
Like nthOneBits(), this one does not need any setup time. It could be used in combination with nthOneBits() to get subsequent patterns after getting an initial one at a given position. nextOneBits() is much faster than nthOneBits(i+1) but is still slower than the generator function.
For very large integers, using nthOneBits() and nextOneBits() may be the only practical options.
You are dealing with permutations of multisets. There are many ways to achieve this and as #BPL points out, doing this efficiently is non-trivial. There are many great methods mentioned here: permutations with unique values. The cleanest (not sure if it's the most efficient), is to use the multiset_permutations from the sympy module.
import time
from sympy.utilities.iterables import multiset_permutations
t = time.process_time()
## Credit to #BPL for the general setup
multiPerms = ["".join(v) for v in multiset_permutations(["1"]*6+["0"]*26)]
elapsed_time = time.process_time() - t
print(elapsed_time)
On my machine, the above computes in just over 8 seconds. It generates just under a million results as well:
len(multiPerms)
906192
I have an arbitrary array with only binary values- say:
a = np.array([1,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0,1,1,0,])
What would be the most efficient way to count average length of sequences of 1s array? Eg. In this example, it would be (1 + 8 + 2)/3.
For an all numpy solution, you can use Alex Martelli's solution like so:
def runs_of_ones_array(bits):
# make sure all runs of ones are well-bounded
bounded = np.hstack(([0], bits, [0]))
# get 1 at run starts and -1 at run ends
difs = np.diff(bounded)
run_starts, = np.where(difs > 0)
run_ends, = np.where(difs < 0)
return run_ends - run_starts
>>> a=np.array([1,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0,1,1,0,])
>>> b=runs_of_ones_array(a)
>>> float(sum(b))/len(b)
3.66666666667
I'm not sure easiest, but one alternative is
np.mean([len(list(v)) for k,v in itertools.groupby(a) if k])
3.6666666666666665
Explanation
groupby groups adjacent equal values together, we filter only ones (if k), the list comprehension [...] creates the list of the lengths of sub sequences of ones, i.e. [1,8,2], and mean computes the average value.
x = [1,2,3,4,5,6,7,8,9,10]
#Random list elements
for i in range(int(len(x)/2)):
value = x[i]
x[i] = x[len(x)-i-1]
x[len(x)-i-1] = value
#Confusion on efficiency
print(x)
This is a uni course for first year. So no python shortcuts are allowed
Not sure what counts as "a shortcut" (reversed and the "Martian Smiley" [::-1] being obvious candidates -- but does either count as "a shortcut"?!), but at least a couple small improvements are easy:
L = len(x)
for i in range(L//2):
mirror = L - i - 1
x[i], x[mirror] = x[mirror], x[i]
This gets len(x) only once -- it's a fast operation but there's no reason to keep repeating it over and over -- also computes mirror but once, does the swap more directly, and halves L (for the range argument) directly with the truncating-division operator rather than using the non-truncating division and then truncating with int. Nanoseconds for each case, but it may be considered slightly clearer as well as microscopically faster.
x = [1,2,3,4,5,6,7,8,9,10]
x = x.__getitem__(slice(None,None,-1))
slice is a python builtin object (like range and len that you used in your example)
__getitem__ is a method belonging to iterable types ( of which x is)
there are absolutely no shortcuts here :) and its effectively one line.
Why am I getting this error, from line 5 of my code, when attempting to solve Project Euler Problem 11?
for x in matrix:
p = 0
for y in x:
if p < 17:
currentProduct = int(y) * int(x[p + 1]) * int(x[p + 2]) * int(x[p + 3])
if currentProduct > highestProduct:
print(currentProduct)
highestProduct = currentProduct
else:
break
p += 1
'generator' object is not subscriptable
Your x value is is a generator object, which is an Iterator: it generates values in order, as they are requested by a for loop or by calling next(x).
You are trying to access it as though it were a list or other Sequence type, which let you access arbitrary elements by index as x[p + 1].
If you want to look up values from your generator's output by index, you may want to convert it to a list:
x = list(x)
This solves your problem, and is suitable in most cases. However, this requires generating and saving all of the values at once, so it can fail if you're dealing with an extremely long or infinite list of values, or the values are extremely large.
If you just needed a single value from the generator, you could instead use itertools.islice(x, p) to discard the first p values, then next(...) to take the one you need. This eliminate the need to hold multiple items in memory or compute values beyond the one you're looking for.
import itertools
result = next(itertools.islice(x, p))
As an extension to Jeremy's answer some thoughts about the design of your code:
Looking at your algorithm, it appears that you do not actually need truly random access to the values produced by the generator: At any point in time you only need to keep four consecutive values (three, with an extra bit of optimization). This is a bit obscured in your code because you mix indexing and iteration: If indexing would work (which it doesn't), your y could be written as x[p + 0].
For such algorithms, you can apply kind of a "sliding window" technique, demonstrated below in a stripped-down version of your code:
import itertools, functools, operator
vs = [int(v) for v in itertools.islice(x, 3)]
for v in x:
vs.append(int(v))
currentProduct = functools.reduce(operator.mul, vs, 1)
print(currentProduct)
vs = vs[1:]
So I was reading the Wikipedia article on the Sieve of Eratosthenes and it included a Python implementation:
http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes#Algorithm_complexity_and_implementation
def eratosthenes_sieve(n):
# Create a candidate list within which non-primes will be
# marked as None; only candidates below sqrt(n) need be checked.
candidates = range(n+1)
fin = int(n**0.5)
# Loop over the candidates, marking out each multiple.
for i in xrange(2, fin+1):
if not candidates[i]:
continue
candidates[2*i::i] = [None] * (n//i - 1)
# Filter out non-primes and return the list.
return [i for i in candidates[2:] if i]
It looks like a very simple and elegant implementation. I've seen other implementations, even in Python, and I understand how the Sieve works. But the particular way this implementation works, I"m getting a little confused. Seems whoever was writing that page was pretty clever.
I get that its iterating through the list, finding primes, and then marking multiples of primes as non-prime.
But what does this line do exactly:
candidates[2*i::i] = [None] * (n//i - 1)
I've figured out that its slicing candidates from 2*i to the end, iterating by i, so that means all multiples of i, start at 2*i, then go to 3*i, then go to 4*i till you finish the list.
But what does [None] * (n//i - 1) mean? Why not just set it to False?
Thanks. Kind of a specific question with a single answer, but I think this is the place to ask it. I would sure appreciate a clear explanation.
candidates[2*i::i] = [None] * (n//i - 1)
is just a terse way of writing
for j in range(2 * i, n, i):
candidates[j] = None
which works by assigning an list of Nones to a slice of candidates.
L * N creates and concatenates N (shallow) copies of L, so [None] * (n//i - 1) gives a list of ceil(n / i) times None. Slice assignment (L[start:end:step] = new_L) overwrites the items of the list the slice touches with the items of new_L.
You are right, one could set the items to False as well - I think this would be preferrable, the author of the code obviously thought None would be a better indicator of "crossed out". But None works as well, as bool(None) is False and .. if i is essentially if bool(i).