This is a collection of exercises that have been collected in the numpy mailing list, on stack overflow and in the numpy documentation. The goal of this collection is to offer a quick reference for both old and new users but also to provide a set of exercises for those who teach.
If you find an error or think you've a better way to solve some of them, feel free to open an issue at https://github.com/rougier/numpy-100
import numpy as npprint(np.__version__)
np.show_config()Z = np.zeros(10)
print(Z)Z = np.zeros((10,10))
print("%d bytes" % (Z.size * Z.itemsize))%run `python -c "import numpy; numpy.info(numpy.add)"`Z = np.zeros(10)
Z[4] = 1
print(Z)Z = np.arange(10,50)
print(Z)Z = np.arange(50)
Z = Z[::-1]
print(Z)Z = np.arange(9).reshape(3,3)
print(Z)nz = np.nonzero([1,2,0,0,4,0])
print(nz)Z = np.eye(3)
print(Z)Z = np.random.random((3,3,3))
print(Z)Z = np.random.random((10,10))
Zmin, Zmax = Z.min(), Z.max()
print(Zmin, Zmax)Z = np.random.random(30)
m = Z.mean()
print(m)Z = np.ones((10,10))
Z[1:-1,1:-1] = 0
print(Z)Z = np.ones((5,5))
Z = np.pad(Z, pad_width=1, mode='constant', constant_values=0)
print(Z)print(0 * np.nan)
print(np.nan == np.nan)
print(np.inf > np.nan)
print(np.nan - np.nan)
print(0.3 == 3 * 0.1)Z = np.diag(1+np.arange(4),k=-1)
print(Z)Z = np.zeros((8,8),dtype=int)
Z[1::2,::2] = 1
Z[::2,1::2] = 1
print(Z)print(np.unravel_index(99,(6,7,8)))Z = np.tile( np.array([[0,1],[1,0]]), (4,4))
print(Z)Z = np.random.random((5,5))
Zmax, Zmin = Z.max(), Z.min()
Z = (Z - Zmin)/(Zmax - Zmin)
print(Z)color = np.dtype([("r", np.ubyte, 1),
("g", np.ubyte, 1),
("b", np.ubyte, 1),
("a", np.ubyte, 1)])Z = np.dot(np.ones((5,3)), np.ones((3,2)))
print(Z)
# Alternative solution, in Python 3.5 and above
Z = np.ones((5,3)) @ np.ones((3,2))
print(Z)# Author: Evgeni Burovski
Z = np.arange(11)
Z[(3 < Z) & (Z <= 8)] *= -1
print(Z)# Author: Jake VanderPlas
print(sum(range(5),-1))
from numpy import *
print(sum(range(5),-1))Z**Z
2 << Z >> 2
Z <- Z
1j*Z
Z/1/1
Z<Z>Zprint(np.array(0) / np.array(0))
print(np.array(0) // np.array(0))
print(np.array([np.nan]).astype(int).astype(float))# Author: Charles R Harris
Z = np.random.uniform(-10,+10,10)
print (np.copysign(np.ceil(np.abs(Z)), Z))Z1 = np.random.randint(0,10,10)
Z2 = np.random.randint(0,10,10)
print(np.intersect1d(Z1,Z2))# Suicide mode on
defaults = np.seterr(all="ignore")
Z = np.ones(1) / 0
# Back to sanity
_ = np.seterr(**defaults)An equivalent way, with a context manager:
with np.errstate(divide='ignore'):
Z = np.ones(1) / 0np.sqrt(-1) == np.emath.sqrt(-1)yesterday = np.datetime64('today', 'D') - np.timedelta64(1, 'D')
today = np.datetime64('today', 'D')
tomorrow = np.datetime64('today', 'D') + np.timedelta64(1, 'D')Z = np.arange('2016-07', '2016-08', dtype='datetime64[D]')
print(Z)A = np.ones(3)*1
B = np.ones(3)*2
C = np.ones(3)*3
np.add(A,B,out=B)
np.divide(A,2,out=A)
np.negative(A,out=A)
np.multiply(A,B,out=A)Z = np.random.uniform(0,10,10)
print (Z - Z%1)
print (np.floor(Z))
print (np.ceil(Z)-1)
print (Z.astype(int))
print (np.trunc(Z))Z = np.zeros((5,5))
Z += np.arange(5)
print(Z)def generate():
for x in range(10):
yield x
Z = np.fromiter(generate(),dtype=float,count=-1)
print(Z)Z = np.linspace(0,1,11,endpoint=False)[1:]
print(Z)Z = np.random.random(10)
Z.sort()
print(Z)# Author: Evgeni Burovski
Z = np.arange(10)
np.add.reduce(Z)A = np.random.randint(0,2,5)
B = np.random.randint(0,2,5)
# Assuming identical shape of the arrays and a tolerance for the comparison of values
equal = np.allclose(A,B)
print(equal)
# Checking both the shape and the element values, no tolerance (values have to be exactly equal)
equal = np.array_equal(A,B)
print(equal)Z = np.zeros(10)
Z.flags.writeable = False
Z[0] = 144. Consider a random 10x2 matrix representing cartesian coordinates, convert them to polar coordinates (★★☆)
Z = np.random.random((10,2))
X,Y = Z[:,0], Z[:,1]
R = np.sqrt(X**2+Y**2)
T = np.arctan2(Y,X)
print(R)
print(T)Z = np.random.random(10)
Z[Z.argmax()] = 0
print(Z)Z = np.zeros((5,5), [('x',float),('y',float)])
Z['x'], Z['y'] = np.meshgrid(np.linspace(0,1,5),
np.linspace(0,1,5))
print(Z)# Author: Evgeni Burovski
X = np.arange(8)
Y = X + 0.5
C = 1.0 / np.subtract.outer(X, Y)
print(np.linalg.det(C))for dtype in [np.int8, np.int32, np.int64]:
print(np.iinfo(dtype).min)
print(np.iinfo(dtype).max)
for dtype in [np.float32, np.float64]:
print(np.finfo(dtype).min)
print(np.finfo(dtype).max)
print(np.finfo(dtype).eps)np.set_printoptions(threshold=np.nan)
Z = np.zeros((16,16))
print(Z)Z = np.arange(100)
v = np.random.uniform(0,100)
index = (np.abs(Z-v)).argmin()
print(Z[index])Z = np.zeros(10, [ ('position', [ ('x', float, 1),
('y', float, 1)]),
('color', [ ('r', float, 1),
('g', float, 1),
('b', float, 1)])])
print(Z)52. Consider a random vector with shape (100,2) representing coordinates, find point by point distances (★★☆)
Z = np.random.random((10,2))
X,Y = np.atleast_2d(Z[:,0], Z[:,1])
D = np.sqrt( (X-X.T)**2 + (Y-Y.T)**2)
print(D)
# Much faster with scipy
import scipy
# Thanks Gavin Heverly-Coulson (#issue 1)
import scipy.spatial
Z = np.random.random((10,2))
D = scipy.spatial.distance.cdist(Z,Z)
print(D)Z = np.arange(10, dtype=np.float32)
Z = Z.astype(np.int32, copy=False)
print(Z)from io import StringIO
# Fake file
s = StringIO("""1, 2, 3, 4, 5\n
6, , , 7, 8\n
, , 9,10,11\n""")
Z = np.genfromtxt(s, delimiter=",", dtype=np.int)
print(Z)Z = np.arange(9).reshape(3,3)
for index, value in np.ndenumerate(Z):
print(index, value)
for index in np.ndindex(Z.shape):
print(index, Z[index])X, Y = np.meshgrid(np.linspace(-1,1,10), np.linspace(-1,1,10))
D = np.sqrt(X*X+Y*Y)
sigma, mu = 1.0, 0.0
G = np.exp(-( (D-mu)**2 / ( 2.0 * sigma**2 ) ) )
print(G)# Author: Divakar
n = 10
p = 3
Z = np.zeros((n,n))
np.put(Z, np.random.choice(range(n*n), p, replace=False),1)
print(Z)# Author: Warren Weckesser
X = np.random.rand(5, 10)
# Recent versions of numpy
Y = X - X.mean(axis=1, keepdims=True)
# Older versions of numpy
Y = X - X.mean(axis=1).reshape(-1, 1)
print(Y)# Author: Steve Tjoa
Z = np.random.randint(0,10,(3,3))
print(Z)
print(Z[Z[:,1].argsort()])# Author: Warren Weckesser
Z = np.random.randint(0,3,(3,10))
print((~Z.any(axis=0)).any())Z = np.random.uniform(0,1,10)
z = 0.5
m = Z.flat[np.abs(Z - z).argmin()]
print(m)62. Considering two arrays with shape (1,3) and (3,1), how to compute their sum using an iterator? (★★☆)
A = np.arange(3).reshape(3,1)
B = np.arange(3).reshape(1,3)
it = np.nditer([A,B,None])
for x,y,z in it: z[...] = x + y
print(it.operands[2])class NamedArray(np.ndarray):
def __new__(cls, array, name="no name"):
obj = np.asarray(array).view(cls)
obj.name = name
return obj
def __array_finalize__(self, obj):
if obj is None: return
self.info = getattr(obj, 'name', "no name")
Z = NamedArray(np.arange(10), "range_10")
print (Z.name)64. Consider a given vector, how to add 1 to each element indexed by a second vector (be careful with repeated indices)? (★★★)
# Author: Brett Olsen
Z = np.ones(10)
I = np.random.randint(0,len(Z),20)
Z += np.bincount(I, minlength=len(Z))
print(Z)
# Another solution
# Author: Bartosz Telenczuk
np.add.at(Z, I, 1)
print(Z)# Author: Alan G Isaac
X = [1,2,3,4,5,6]
I = [1,3,9,3,4,1]
F = np.bincount(I,X)
print(F)# Author: Nadav Horesh
w,h = 16,16
I = np.random.randint(0,2,(h,w,3)).astype(np.ubyte)
F = I[...,0]*256*256 + I[...,1]*256 +I[...,2]
n = len(np.unique(F))
print(np.unique(I))A = np.random.randint(0,10,(3,4,3,4))
# solution by passing a tuple of axes (introduced in numpy 1.7.0)
sum = A.sum(axis=(-2,-1))
print(sum)
# solution by flattening the last two dimensions into one
# (useful for functions that don't accept tuples for axis argument)
sum = A.reshape(A.shape[:-2] + (-1,)).sum(axis=-1)
print(sum)68. Considering a one-dimensional vector D, how to compute means of subsets of D using a vector S of same size describing subset indices? (★★★)
# Author: Jaime Fernández del Río
D = np.random.uniform(0,1,100)
S = np.random.randint(0,10,100)
D_sums = np.bincount(S, weights=D)
D_counts = np.bincount(S)
D_means = D_sums / D_counts
print(D_means)
# Pandas solution as a reference due to more intuitive code
import pandas as pd
print(pd.Series(D).groupby(S).mean())# Author: Mathieu Blondel
A = np.random.uniform(0,1,(5,5))
B = np.random.uniform(0,1,(5,5))
# Slow version
np.diag(np.dot(A, B))
# Fast version
np.sum(A * B.T, axis=1)
# Faster version
np.einsum("ij,ji->i", A, B)70. Consider the vector [1, 2, 3, 4, 5], how to build a new vector with 3 consecutive zeros interleaved between each value? (★★★)
# Author: Warren Weckesser
Z = np.array([1,2,3,4,5])
nz = 3
Z0 = np.zeros(len(Z) + (len(Z)-1)*(nz))
Z0[::nz+1] = Z
print(Z0)71. Consider an array of dimension (5,5,3), how to mulitply it by an array with dimensions (5,5)? (★★★)
A = np.ones((5,5,3))
B = 2*np.ones((5,5))
print(A * B[:,:,None])# Author: Eelco Hoogendoorn
A = np.arange(25).reshape(5,5)
A[[0,1]] = A[[1,0]]
print(A)73. Consider a set of 10 triplets describing 10 triangles (with shared vertices), find the set of unique line segments composing all the triangles (★★★)
# Author: Nicolas P. Rougier
faces = np.random.randint(0,100,(10,3))
F = np.roll(faces.repeat(2,axis=1),-1,axis=1)
F = F.reshape(len(F)*3,2)
F = np.sort(F,axis=1)
G = F.view( dtype=[('p0',F.dtype),('p1',F.dtype)] )
G = np.unique(G)
print(G)74. Given an array C that is a bincount, how to produce an array A such that np.bincount(A) == C? (★★★)
# Author: Jaime Fernández del Río
C = np.bincount([1,1,2,3,4,4,6])
A = np.repeat(np.arange(len(C)), C)
print(A)# Author: Jaime Fernández del Río
def moving_average(a, n=3) :
ret = np.cumsum(a, dtype=float)
ret[n:] = ret[n:] - ret[:-n]
return ret[n - 1:] / n
Z = np.arange(20)
print(moving_average(Z, n=3))76. Consider a one-dimensional array Z, build a two-dimensional array whose first row is (Z[0],Z[1],Z[2]) and each subsequent row is shifted by 1 (last row should be (Z[-3],Z[-2],Z[-1]) (★★★)
# Author: Joe Kington / Erik Rigtorp
from numpy.lib import stride_tricks
def rolling(a, window):
shape = (a.size - window + 1, window)
strides = (a.itemsize, a.itemsize)
return stride_tricks.as_strided(a, shape=shape, strides=strides)
Z = rolling(np.arange(10), 3)
print(Z)# Author: Nathaniel J. Smith
Z = np.random.randint(0,2,100)
np.logical_not(Z, out=Z)
Z = np.random.uniform(-1.0,1.0,100)
np.negative(Z, out=Z)78. Consider 2 sets of points P0,P1 describing lines (2d) and a point p, how to compute distance from p to each line i (P0[i],P1[i])? (★★★)
def distance(P0, P1, p):
T = P1 - P0
L = (T**2).sum(axis=1)
U = -((P0[:,0]-p[...,0])*T[:,0] + (P0[:,1]-p[...,1])*T[:,1]) / L
U = U.reshape(len(U),1)
D = P0 + U*T - p
return np.sqrt((D**2).sum(axis=1))
P0 = np.random.uniform(-10,10,(10,2))
P1 = np.random.uniform(-10,10,(10,2))
p = np.random.uniform(-10,10,( 1,2))
print(distance(P0, P1, p))79. Consider 2 sets of points P0,P1 describing lines (2d) and a set of points P, how to compute distance from each point j (P[j]) to each line i (P0[i],P1[i])? (★★★)
# Author: Italmassov Kuanysh
# based on distance function from previous question
P0 = np.random.uniform(-10, 10, (10,2))
P1 = np.random.uniform(-10,10,(10,2))
p = np.random.uniform(-10, 10, (10,2))
print(np.array([distance(P0,P1,p_i) for p_i in p]))80. Consider an arbitrary array, write a function that extract a subpart with a fixed shape and centered on a given element (pad with a fill value when necessary) (★★★)
# Author: Nicolas Rougier
Z = np.random.randint(0,10,(10,10))
shape = (5,5)
fill = 0
position = (1,1)
R = np.ones(shape, dtype=Z.dtype)*fill
P = np.array(list(position)).astype(int)
Rs = np.array(list(R.shape)).astype(int)
Zs = np.array(list(Z.shape)).astype(int)
R_start = np.zeros((len(shape),)).astype(int)
R_stop = np.array(list(shape)).astype(int)
Z_start = (P-Rs//2)
Z_stop = (P+Rs//2)+Rs%2
R_start = (R_start - np.minimum(Z_start,0)).tolist()
Z_start = (np.maximum(Z_start,0)).tolist()
R_stop = np.maximum(R_start, (R_stop - np.maximum(Z_stop-Zs,0))).tolist()
Z_stop = (np.minimum(Z_stop,Zs)).tolist()
r = [slice(start,stop) for start,stop in zip(R_start,R_stop)]
z = [slice(start,stop) for start,stop in zip(Z_start,Z_stop)]
R[r] = Z[z]
print(Z)
print(R)81. Consider an array Z = [1,2,3,4,5,6,7,8,9,10,11,12,13,14], how to generate an array R = [[1,2,3,4], [2,3,4,5], [3,4,5,6], ..., [11,12,13,14]]? (★★★)
# Author: Stefan van der Walt
Z = np.arange(1,15,dtype=np.uint32)
R = stride_tricks.as_strided(Z,(11,4),(4,4))
print(R)# Author: Stefan van der Walt
Z = np.random.uniform(0,1,(10,10))
U, S, V = np.linalg.svd(Z) # Singular Value Decomposition
rank = np.sum(S > 1e-10)
print(rank)Z = np.random.randint(0,10,50)
print(np.bincount(Z).argmax())# Author: Chris Barker
Z = np.random.randint(0,5,(10,10))
n = 3
i = 1 + (Z.shape[0]-3)
j = 1 + (Z.shape[1]-3)
C = stride_tricks.as_strided(Z, shape=(i, j, n, n), strides=Z.strides + Z.strides)
print(C)# Author: Eric O. Lebigot
# Note: only works for 2d array and value setting using indices
class Symetric(np.ndarray):
def __setitem__(self, index, value):
i,j = index
super(Symetric, self).__setitem__((i,j), value)
super(Symetric, self).__setitem__((j,i), value)
def symetric(Z):
return np.asarray(Z + Z.T - np.diag(Z.diagonal())).view(Symetric)
S = symetric(np.random.randint(0,10,(5,5)))
S[2,3] = 42
print(S)86. Consider a set of p matrices wich shape (n,n) and a set of p vectors with shape (n,1). How to compute the sum of of the p matrix products at once? (result has shape (n,1)) (★★★)
# Author: Stefan van der Walt
p, n = 10, 20
M = np.ones((p,n,n))
V = np.ones((p,n,1))
S = np.tensordot(M, V, axes=[[0, 2], [0, 1]])
print(S)
# It works, because:
# M is (p,n,n)
# V is (p,n,1)
# Thus, summing over the paired axes 0 and 0 (of M and V independently),
# and 2 and 1, to remain with a (n,1) vector.# Author: Robert Kern
Z = np.ones((16,16))
k = 4
S = np.add.reduceat(np.add.reduceat(Z, np.arange(0, Z.shape[0], k), axis=0),
np.arange(0, Z.shape[1], k), axis=1)
print(S)# Author: Nicolas Rougier
def iterate(Z):
# Count neighbours
N = (Z[0:-2,0:-2] + Z[0:-2,1:-1] + Z[0:-2,2:] +
Z[1:-1,0:-2] + Z[1:-1,2:] +
Z[2: ,0:-2] + Z[2: ,1:-1] + Z[2: ,2:])
# Apply rules
birth = (N==3) & (Z[1:-1,1:-1]==0)
survive = ((N==2) | (N==3)) & (Z[1:-1,1:-1]==1)
Z[...] = 0
Z[1:-1,1:-1][birth | survive] = 1
return Z
Z = np.random.randint(0,2,(50,50))
for i in range(100): Z = iterate(Z)
print(Z)Z = np.arange(10000)
np.random.shuffle(Z)
n = 5
# Slow
print (Z[np.argsort(Z)[-n:]])
# Fast
print (Z[np.argpartition(-Z,n)[:n]])90. Given an arbitrary number of vectors, build the cartesian product (every combinations of every item) (★★★)
# Author: Stefan Van der Walt
def cartesian(arrays):
arrays = [np.asarray(a) for a in arrays]
shape = (len(x) for x in arrays)
ix = np.indices(shape, dtype=int)
ix = ix.reshape(len(arrays), -1).T
for n, arr in enumerate(arrays):
ix[:, n] = arrays[n][ix[:, n]]
return ix
print (cartesian(([1, 2, 3], [4, 5], [6, 7])))Z = np.array([("Hello", 2.5, 3),
("World", 3.6, 2)])
R = np.core.records.fromarrays(Z.T,
names='col1, col2, col3',
formats = 'S8, f8, i8')
print(R)# Author: Ryan G.
x = np.random.rand(5e7)
%timeit np.power(x,3)
%timeit x*x*x
%timeit np.einsum('i,i,i->i',x,x,x)93. Consider two arrays A and B of shape (8,3) and (2,2). How to find rows of A that contain elements of each row of B regardless of the order of the elements in B? (★★★)
# Author: Gabe Schwartz
A = np.random.randint(0,5,(8,3))
B = np.random.randint(0,5,(2,2))
C = (A[..., np.newaxis, np.newaxis] == B)
rows = np.where(C.any((3,1)).all(1))[0]
print(rows)# Author: Robert Kern
Z = np.random.randint(0,5,(10,3))
print(Z)
# solution for arrays of all dtypes (including string arrays and record arrays)
E = np.all(Z[:,1:] == Z[:,:-1], axis=1)
U = Z[~E]
print(U)
# soluiton for numerical arrays only, will work for any number of columns in Z
U = Z[Z.max(axis=1) != Z.min(axis=1),:]
print(U)# Author: Warren Weckesser
I = np.array([0, 1, 2, 3, 15, 16, 32, 64, 128])
B = ((I.reshape(-1,1) & (2**np.arange(8))) != 0).astype(int)
print(B[:,::-1])
# Author: Daniel T. McDonald
I = np.array([0, 1, 2, 3, 15, 16, 32, 64, 128], dtype=np.uint8)
print(np.unpackbits(I[:, np.newaxis], axis=1))# Author: Jaime Fernández del Río
Z = np.random.randint(0,2,(6,3))
T = np.ascontiguousarray(Z).view(np.dtype((np.void, Z.dtype.itemsize * Z.shape[1])))
_, idx = np.unique(T, return_index=True)
uZ = Z[idx]
print(uZ)97. Considering 2 vectors A & B, write the einsum equivalent of inner, outer, sum, and mul function (★★★)
# Author: Alex Riley
# Make sure to read: http://ajcr.net/Basic-guide-to-einsum/
A = np.random.uniform(0,1,10)
B = np.random.uniform(0,1,10)
np.einsum('i->', A) # np.sum(A)
np.einsum('i,i->i', A, B) # A * B
np.einsum('i,i', A, B) # np.inner(A, B)
np.einsum('i,j->ij', A, B) # np.outer(A, B)98. Considering a path described by two vectors (X,Y), how to sample it using equidistant samples (★★★)?
# Author: Bas Swinckels
phi = np.arange(0, 10*np.pi, 0.1)
a = 1
x = a*phi*np.cos(phi)
y = a*phi*np.sin(phi)
dr = (np.diff(x)**2 + np.diff(y)**2)**.5 # segment lengths
r = np.zeros_like(x)
r[1:] = np.cumsum(dr) # integrate path
r_int = np.linspace(0, r.max(), 200) # regular spaced path
x_int = np.interp(r_int, r, x) # integrate path
y_int = np.interp(r_int, r, y)99. Given an integer n and a 2D array X, select from X the rows which can be interpreted as draws from a multinomial distribution with n degrees, i.e., the rows which only contain integers and which sum to n. (★★★)
# Author: Evgeni Burovski
X = np.asarray([[1.0, 0.0, 3.0, 8.0],
[2.0, 0.0, 1.0, 1.0],
[1.5, 2.5, 1.0, 0.0]])
n = 4
M = np.logical_and.reduce(np.mod(X, 1) == 0, axis=-1)
M &= (X.sum(axis=-1) == n)
print(X[M])100. Compute bootstrapped 95% confidence intervals for the mean of a 1D array X (i.e., resample the elements of an array with replacement N times, compute the mean of each sample, and then compute percentiles over the means). (★★★)
# Author: Jessica B. Hamrick
X = np.random.randn(100) # random 1D array
N = 1000 # number of bootstrap samples
idx = np.random.randint(0, X.size, (N, X.size))
means = X[idx].mean(axis=1)
confint = np.percentile(means, [2.5, 97.5])
print(confint)