def prob3():
#problem 3
print "problem 3"
A = rand(1000, 1000)
number = 100
tm = ti("A.reshape(A.size)", setup="from __main__ import A", number=number)
print "A.reshape(A.size) executed ", number, " times in ", tm, " seconds."
number = 100
tm = ti("A.flatten()", setup="from __main__ import A", number=number)
print "A.flatten() executed ", number, " times in ", tm, " seconds."
number = ti("A.reshape((1,A.size))",
setup="from __main__ import A",
number=number)
print "A.reshape((1,A.size)) executed ", number, " times in ", tm, " seconds."
#part 2 of problem 3
print "the difference is that np.vstack(A) returns a new array"
print "while A.T returns a view"
print "A.T is much faster"
A = rand(1, 1000000)
number = 500
tm = ti("np.vstack(A)",
setup="import numpy as np; from __main__ import A",
number=number)
print "np.vstack(A) executed ", number, " times in ", tm, " seconds."
number = 500
tm = ti("A.T", setup="from __main__ import A", number=number)
print "A.T executed ", number, " times in ", tm, " seconds."
def time(sizes):
t1 = []
t2 = []
t3 = []
t4 = []
for size in sizes:
A = rand(size, 10)
B = rand(size, 10)
t1.append(
ti("dist1(A, B)",
setup="from __main__ import dist1, A, B",
number=10))
t2.append(
ti("dist2(A, B)",
setup="from __main__ import dist2, A, B",
number=10))
t3.append(
ti("dist3(A, B)",
setup="from __main__ import dist3, A, B",
number=10))
t4.append(
ti("dist4(A, B)",
setup="from __main__ import dist4, A, B",
number=10))
print size
times = np.empty((5, len(sizes)))
times[0] = sizes
times[1] = t1
times[2] = t2
times[3] = t3
times[4] = t4
np.save("times.npy", times)
def ltest2():
start = ti(number=1000)
l = []
for i in range(1000):
l.append(i)
end = ti(number=1000)
return start - end
def ltest1():
start = ti(number=1000)
l = []
for i in range(1000):
l = l + [i]
end = ti(number=1000)
return start - end
def RepeatingEvenList(aList):
s = ti(number=10000)
res = 0
for i in aList:
res ^= i
e = ti(number=10000)
return res, s - e
def prob1():
k = 200
A = [range(i, i+k) for i in range(0, k**2, k)]
number = 5
print "problem 1"
tm = ti("arrmul(A,A)", setup="from __main__ import A, arrmul", number=number)
print "arrmul(A,A) executed ", number, " times in ", tm, " seconds."
A = np.array(A)
number = 5
tm = ti("np.dot(A,A)", setup="import numpy as np; from __main__ import A", number=number)
print "np.dot(A,A) executed ", number, " times in ", tm, " seconds."
def array_vs_list():
k = 200
A = [range(i, i+k) for i in range(0, k**2, k)]
number = 5
print "problem 1"
tm = ti("arrmul(A,A)", setup="from __main__ import A, arrmul", number=number)
print "arrmul(A,A) executed ", number, " times in ", tm, " seconds."
A = np.array(A)
number = 5
tm = ti("np.dot(A,A)", setup="import numpy as np; from __main__ import A", number=number)
print "np.dot(A,A) executed ", number, " times in ", tm, " seconds."
def time(sizes):
t1 = []
t2 = []
t3 = []
t4 = []
for size in sizes:
A = rand(size, 10)
B = rand(size, 10)
t1.append(ti("dist1(A, B)", setup="from __main__ import dist1, A, B", number=10))
t2.append(ti("dist2(A, B)", setup="from __main__ import dist2, A, B", number=10))
t3.append(ti("dist3(A, B)", setup="from __main__ import dist3, A, B", number=10))
t4.append(ti("dist4(A, B)", setup="from __main__ import dist4, A, B", number=10))
print size
times = np.empty((5, len(sizes)))
times[0] = sizes
times[1] = t1
times[2] = t2
times[3] = t3
times[4] = t4
np.save("times.npy", times)
def UsingDict(aList):
s = ti(number=10000)
#d={i: aList.count(i) for i in aList }
d = {}
for i in aList:
if i in d:
d[i] += 1
else:
d[i] = 1
res = 0
for k, v in d.items():
if v == 1:
res = k
e = ti(number=10000)
return res, s - e
def prob3():
A = rand(1000,1000)
number = 100
tm = ti("A.reshape(A.size)", setup="from __main__ import A", number=number)
print "A.reshape(A.size) executed ", number, " times in ", tm, " seconds."
number = 100
tm = ti("A.flatten()", setup="from __main__ import A", number=number)
print "A.flatten() executed ", number, " times in ", tm, " seconds."
number = ti("A.reshape((1,A.size))", setup="from __main__ import A", number=number)
print "A.reshape((1,A.size)) executed ", number, " times in ", tm, " seconds."
#part 2 of problem 3
print "the difference is that np.vstack(A) returns a new array"
print "while A.T returns a view"
print "A.T is much faster"
A = rand(1,1000000)
number = 500
tm = ti("np.vstack(A)", setup="import numpy as np; from __main__ import A", number=number)
print "np.vstack(A) executed ", number, " times in ", tm, " seconds."
number = 500
tm = ti("A.T", setup="from __main__ import A", number=number)
print "A.T executed ", number, " times in ", tm, " seconds."
from kivy.uix.widget import Widget from random import randint import timeit.timeit as ti from functools import partial import utilfuncs import circle_collide as cc if __name__ == '__main__': t1 = #[Widget(pos=(randint(0, 800), randint(0, 600)) #for i in range(2)] ti(partial(cc, t1[0], t1[1])) ti(t1[0].collide_widget())
i = 2
num = [0] * (n + 1)
p = []
while (i * i <= n):
j = i
while (j * i <= n):
num[i * j] = 1
j += 1
i += 1
for i in xrange(2, n + 1):
if num[i] == 0:
p.append(i)
return p
def factorize(n):
sq = int(n**0.5)
p = seive(sq)
c = Counter()
for prime in p:
while (n % prime == 0):
n /= prime
c[prime] += 1
return c
#print factorize(10000)
print ti("factorize(1213231313131)",
setup="from __main__ import factorize",
number=1)
def ltest4():
start = ti(number=1000)
l = list(range(1000))
end = ti(number=1000)
return start - end
def smaller_than_3(x):
if x < 3:
return True
else:
return False
grp = groupby(a, smaller_than_3)
print("groupby fun:\n")
#print(list(grp))
for key, value in grp:
print(key, list(value))
count = 0
start = ti()
for i in cycle(a):
print(i)
count += 1
if count == 10:
break
end = ti()
print("time taken for cycle fun: " + str(end - start))
start = ti()
for i in repeat(5, 10):
print(i)
end = ti()
print("time taken for cycle fun: " + str(end - start))
def ltest3():
start = ti(number=1000)
l = [i for i in range(1000)]
end = ti(number=1000)
return start - end
"""
Lattice paths
Submit
Show HTML problem content
Problem 15
Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.
How many such routes are there through a 20×20 grid?
Answer: 137846528820
Completed on Sun, 8 Jul 2018, 15:46
"""
import math
from timeit import default_timer as ti
import os.path as op
# Solution 1
def latticePaths(n, m):
return int(math.factorial(n+m) / (math.factorial(n)*math.factorial(m)))
t11 = ti()
print(f"The answer to Project Euler, {op.basename(__file__)[:-3]} is: " + str(latticePaths(20, 20)))
t12 = ti()
print("Time lapsed for solution number 1: " + str((t12-t11)*10**6) + u'\u03BC' + "s")
50000, 100000, 500000, 1000000,
5000000, 10000000, 15000000, 20000000,
25000000])
rowdot_sizes = np.array([10, 50, 100, 200, 300, 400, 500,
600, 700, 800, 900, 1000, 1100,
1200, 1300, 1400, 1500, 1600])
dot_times = np.empty((len(dot_sizes),3))
rowdot_times = np.empty((len(rowdot_sizes),3))
for n, times in zip(dot_sizes, dot_times):
A = rand(n)
B = rand(n)
times[0] = ti("pydot(A, B)", setup="from __main__ import A, B, pydot", number=5) / 5.
times[1] = ti("cydot(A, B)", setup="from __main__ import A, B, cydot", number=500) / 500.
times[2] = ti("A.dot(B)", setup="from __main__ import A, B", number = 500) / 500.
for n, times in zip(rowdot_sizes, rowdot_times):
A = rand(n, 3)
times[0] = ti("pyrowdot(A)", setup="from __main__ import A, pyrowdot", number=1)
times[1] = ti("cyrowdot(A)", setup="from __main__ import A, cyrowdot", number=10) / 10.
times[2] = ti("A.dot(A.T)", setup="from __main__ import A", number=10) / 10.
# plot dot product results
ax = plt.subplot(1,1,1)
p1, = plt.plot(dot_sizes, np.log(dot_times[:,0]), label="Python times")
p2, = plt.plot(dot_sizes, np.log(dot_times[:,1]), label="Cython times")
p3, = plt.plot(dot_sizes, np.log(dot_times[:,2]), label="Numpy with MKL times")
handles, labels = ax.get_legend_handles_labels()
10000000, 15000000, 20000000, 25000000
])
rowdot_sizes = np.array([
10, 50, 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 1100, 1200,
1300, 1400, 1500, 1600
])
dot_times = np.empty((len(dot_sizes), 3))
rowdot_times = np.empty((len(rowdot_sizes), 3))
for n, times in zip(dot_sizes, dot_times):
A = rand(n)
B = rand(n)
times[0] = ti(
"pydot(A, B)", setup="from __main__ import A, B, pydot", number=5) / 5.
times[1] = ti("cydot(A, B)",
setup="from __main__ import A, B, cydot",
number=500) / 500.
times[2] = ti("A.dot(B)", setup="from __main__ import A, B",
number=500) / 500.
for n, times in zip(rowdot_sizes, rowdot_times):
A = rand(n, 3)
times[0] = ti("pyrowdot(A)",
setup="from __main__ import A, pyrowdot",
number=1)
times[1] = ti("cyrowdot(A)",
setup="from __main__ import A, cyrowdot",
number=10) / 10.
times[2] = ti("A.dot(A.T)", setup="from __main__ import A",
from timeit import default_timer as ti my_list = ['abc'] * 10 start = ti() my_string = ''.join(my_list) stop = ti() print(stop - start) print(my_string)
