本文整理汇总了Python中scipy.misc.comb函数的典型用法代码示例。如果您正苦于以下问题:Python comb函数的具体用法?Python comb怎么用?Python comb使用的例子?那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。
在下文中一共展示了comb函数的20个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于我们的系统推荐出更棒的Python代码示例。
示例1: test_cross_validator_with_default_indices
def test_cross_validator_with_default_indices():
n_samples = 4
n_unique_labels = 4
n_folds = 2
p = 2
n_iter = 10 # (the default value)
X = np.array([[1, 2], [3, 4], [5, 6], [7, 8]])
X_1d = np.array([1, 2, 3, 4])
y = np.array([1, 1, 2, 2])
labels = np.array([1, 2, 3, 4])
loo = LeaveOneOut()
lpo = LeavePOut(p)
kf = KFold(n_folds)
skf = StratifiedKFold(n_folds)
lolo = LeaveOneLabelOut()
lopo = LeavePLabelOut(p)
ss = ShuffleSplit(random_state=0)
ps = PredefinedSplit([1, 1, 2, 2]) # n_splits = np of unique folds = 2
n_splits = [n_samples, comb(n_samples, p), n_folds, n_folds,
n_unique_labels, comb(n_unique_labels, p), n_iter, 2]
for i, cv in enumerate([loo, lpo, kf, skf, lolo, lopo, ss, ps]):
# Test if get_n_splits works correctly
assert_equal(n_splits[i], cv.get_n_splits(X, y, labels))
# Test if the cross-validator works as expected even if
# the data is 1d
np.testing.assert_equal(list(cv.split(X, y, labels)),
list(cv.split(X_1d, y, labels)))
# Test that train, test indices returned are integers
for train, test in cv.split(X, y, labels):
assert_equal(np.asarray(train).dtype.kind, 'i')
assert_equal(np.asarray(train).dtype.kind, 'i')
开发者ID:absolutelyNoWarranty,项目名称:scikit-learn,代码行数:35,代码来源:test_split.py
示例2: findF
def findF(k,p,q,x, side):
#print k, p, q, x
if k in Fdict:
if p in Fdict[k]:
if q in Fdict[k][p]:
if x in Fdict[k][p][q]:
if side in Fdict[k][p][q][x]:
return Fdict[k][p][q][x][side]
if q>0:
if side==LEFT: #left
ret=0
for l in xrange(min(p*q+1, x+1)):
ret+=findF(k, p, 0, x-l, LEFT)*sc.comb(p*q,l)
elif side==RIGHT: #right
ret=0
for l in xrange(min(p*q+1, x+1)):
if l==0:
ret+=0 #findF(k, p, 0, x-l, LEFT) it requires at least one edge from left to right.
else:
ret+=findF(k, p, 0, x-l, LEFT)*sc.comb(q,l) #external nodes can only connect the "root" of the right tree
#ret+=findF(k, p, 0, x-l, LEFT)*sc.comb(q,1)*sc.comb((p-1)*q,l-1)
else:
if k==0:
if p==1:
if q==0:
if x==0:
ret=1
else:
ret=0
else:
print "Error 1"
else:
print "Errro 0"
else:
ret=0
for i in xrange(k):
tmp=0
for j in xrange(k-i-1,x-i):
tmp += findF(k-i-1,k-i,i+1,j, LEFT)*findF(i, i+1, k-i, x-j, RIGHT)
ret += tmp*sc.comb(k-1,i)
if k not in Fdict:
Fdict[k]={}
if p not in Fdict[k]:
Fdict[k][p]={}
if q not in Fdict[k][p]:
Fdict[k][p][q]={}
if x not in Fdict[k][p][q]:
Fdict[k][p][q][x]={}
Fdict[k][p][q][x][side]=ret
return ret
开发者ID:simcbr,项目名称:misc,代码行数:60,代码来源:verifyF2.py
示例3: brive
def brive(N, replace_zeros=True):
"""
The brive estimator
Parameters
----------
N : np.array, int
Counts vector
replace_zeros: bool
Replaces zeros with uniform prior
Returns
-------
pvals
"""
N = N.astype(np.int)
n = sum(N)
pvals = np.zeros(len(N), dtype=np.float64)
for i in range(len(N)):
if N[i]==0 or N[i]==1: continue
trials = [comb(t-1, N[i]-1) / (t * (comb(n, N[i])))
for t in range(N[i], n+1)]
pvals[i] = (float(N[i]-1)) * sum(trials)
if replace_zeros:
m = sum(pvals)
if 0 < m < 1 and (pvals==0).sum() > 0:
pvals[pvals==0] = (1 - m) / (pvals==0).sum()
return pvals
开发者ID:mortonjt,项目名称:benchmarkThis,代码行数:29,代码来源:_coverage.py
示例4: solution3
def solution3():
'''
>>> solution3()
137846528820
'''
from scipy.misc import comb
print comb(40, 20, exact=True)
开发者ID:youess,项目名称:project_euler_python,代码行数:7,代码来源:015.py
示例5: rare
def rare(y, size):
notabs = ~np.isnan(y)
t = y[notabs]
N = np.sum(t)
diff = N - t
rare_calc = np.sum(1 - comb(diff, size)/comb(N, size))
return rare_calc
开发者ID:grovduck,项目名称:ecopy,代码行数:7,代码来源:rarefy.py
示例6: eval_gof
def eval_gof( self, data):
probs = self.eval(data.levels)
expected = probs * data.Ntrials
# Compute various goodness of fit statistics
LL = -2*sum((data.Ncorr*np.log(expected/(data.Ncorr+np.finfo(float).eps))
+(data.Ntrials-data.Ncorr)*np.log((data.Ntrials-expected)/(data.Ntrials-data.Ncorr+np.finfo(float).eps))));
X2 = sum((data.Ncorr-expected)**2./expected/(1.-probs))
# Adding eps to avoid log(0) Nans
self.prinsNLL = -sum( data.Ncorr*np.log(probs+np.finfo(float).eps)+(data.Ntrials-data.Ncorr)*np.log(1.-probs+np.finfo(float).eps) )
# Treutwein/Strasburger 1999 Eq 6 (likelihood of the data)
L_ts = 2**(sum( data.Ntrials ))
LL_ts = 0.0
#L_ts = 1.0
for level in np.arange( len(data.levels) ):
# TODO: is right to use observed data or function values?: next two lines can chg to try fitted
thisN = data.Ntrials[level]
thisCorr = data.Ncorr[level]
L_ts *= misc.comb( thisN, thisCorr ) * (probs[level]**thisCorr) * (1.0 - probs[level])**(thisN-thisCorr)
LL_ts += np.log(misc.comb( thisN, thisCorr )) + thisCorr*np.log(probs[level]) +np.log(1.0 - probs[level])*(thisN-thisCorr)
#TODO: This is how Prins' clamps the lapse. Parameterize.
if (self.params[self.PARAM_UPPER] < 0) or (self.params[self.PARAM_UPPER] > 0.05):
self.prinsNLL=np.inf
return probs,LL,X2,L_ts,LL_ts,self.prinsNLL
开发者ID:ivanov,项目名称:PyPsy,代码行数:27,代码来源:pf.py
示例7: pdf
def pdf(self, x, k, n, p):
'''distribution of success runs of length k or more
Parameters
----------
x : float
count of runs of length n
k : int
length of runs
n : int
total number of observations or trials
p : float
probability of success in each Bernoulli trial
Returns
-------
pdf : float
probability that x runs of length of k are observed
Notes
-----
not yet vectorized
References
----------
Muselli 1996, theorem 3
'''
q = 1-p
m = np.arange(x, (n+1)//(k+1)+1)[:,None]
terms = (-1)**(m-x) * comb(m, x) * p**(m*k) * q**(m-1) \
* (comb(n - m*k, m - 1) + q * comb(n - m*k, m))
return terms.sum(0)
开发者ID:0ceangypsy,项目名称:statsmodels,代码行数:33,代码来源:runs.py
示例8: daub
def daub(p):
"""
The coefficients for the FIR low-pass filter producing Daubechies wavelets.
p>=1 gives the order of the zero at f=1/2.
There are 2p filter coefficients.
Parameters
----------
p : int
Order of the zero at f=1/2, can have values from 1 to 34.
"""
sqrt = np.sqrt
if p < 1:
raise ValueError("p must be at least 1.")
if p==1:
c = 1/sqrt(2)
return np.array([c,c])
elif p==2:
f = sqrt(2)/8
c = sqrt(3)
return f*np.array([1+c,3+c,3-c,1-c])
elif p==3:
tmp = 12*sqrt(10)
z1 = 1.5 + sqrt(15+tmp)/6 - 1j*(sqrt(15)+sqrt(tmp-15))/6
z1c = np.conj(z1)
f = sqrt(2)/8
d0 = np.real((1-z1)*(1-z1c))
a0 = np.real(z1*z1c)
a1 = 2*np.real(z1)
return f/d0*np.array([a0, 3*a0-a1, 3*a0-3*a1+1, a0-3*a1+3, 3-a1, 1])
elif p<35:
# construct polynomial and factor it
if p<35:
P = [comb(p-1+k,k,exact=1) for k in range(p)][::-1]
yj = np.roots(P)
else: # try different polynomial --- needs work
P = [comb(p-1+k,k,exact=1)/4.0**k for k in range(p)][::-1]
yj = np.roots(P) / 4
# for each root, compute two z roots, select the one with |z|>1
# Build up final polynomial
c = np.poly1d([1,1])**p
q = np.poly1d([1])
for k in range(p-1):
yval = yj[k]
part = 2*sqrt(yval*(yval-1))
const = 1-2*yval
z1 = const + part
if (abs(z1)) < 1:
z1 = const - part
q = q * [1,-z1]
q = c * np.real(q)
# Normalize result
q = q / np.sum(q) * sqrt(2)
return q.c[::-1]
else:
raise ValueError("Polynomial factorization does not work "
"well for p too large.")
开发者ID:258073127,项目名称:MissionPlanner,代码行数:60,代码来源:wavelets.py
示例9: ARI
def ARI(trueLab, predLab):
"""
compute adjusted rand index, ranges in [-1, 1] with random assignment score 0 and perfect score 1
:param trueLab: ground truth labels
:param predLab: predicted labels
:return: adjusted rand index
"""
n = len(trueLab)
trueLab = np.array(trueLab)
predLab = np.array(predLab)
trueCluster = dict(zip(set(trueLab), [np.where(trueLab == x)[0] for x in set(trueLab)]))
predCluster = dict(zip(set(predLab), [np.where(predLab == x)[0] for x in set(predLab)]))
nTrue = len(trueCluster)
nPred = len(predCluster)
cTable = np.zeros((nTrue, nPred))
for i in range(nTrue):
for j in range(nPred):
cTable[i, j] = len(np.intersect1d(trueCluster.values()[i], predCluster.values()[j]))
a = comb(np.sum(cTable, axis=1), 2).sum()
b = comb(np.sum(cTable, axis=0), 2).sum()
c = comb(n, 2)
return (comb(cTable, 2).sum() - (a * b) / c) / (0.5 * (a + b) - (a * b) / c)
开发者ID:DennisLZL,项目名称:ari,代码行数:28,代码来源:ARI.py
示例10: get_motifspace_size
def get_motifspace_size(q, n):
"""return length of search space according to equation which is mentioned in Section 3.1 of the paper"""
return reduce(
lambda x, y: x + (int(sc.comb(q, y, exact=True)) * 4 ** (q - y)),
[i for i in range(1, n + 1)],
int(sc.comb(q, 0, exact=True)) * 4 ** (q - 0),
)
开发者ID:gatoravi,项目名称:nowarranty,代码行数:7,代码来源:discovering_cse_v2.py
示例11: estimate
def estimate(E, r, f, lp):
if f <= r:
return 0.0
if E - f < lp:
return 1.0
probs = [(1 - (1.0 * scal(comb(E - f, lp)) / scal(comb(e, lp)))) for e in range(E, E - (r+ 1), -1)]
return reduce(lambda x, y: x * y, probs, 1.0)
开发者ID:bestephe,项目名称:res-sim,代码行数:7,代码来源:bound_res.py
示例12: bilinear
def bilinear(b, a, fs=1.0):
"""Return a digital filter from an analog filter using the bilinear transform.
The bilinear transform substitutes ``(z-1) / (z+1``) for ``s``.
"""
#This function has been copied out of scipy
fs =float(fs)
a,b = map(num.atleast_1d,(a,b))
D = len(a) - 1
N = len(b) - 1
artype = float
M = max([N,D])
Np = M
Dp = M
bprime = num.zeros(Np+1,artype)
aprime = num.zeros(Dp+1,artype)
for j in range(Np+1):
val = 0.0
for i in range(N+1):
for k in range(i+1):
for l in range(M-i+1):
if k+l == j:
val += comb(i,k)*comb(M-i,l)*b[N-i]*pow(2*fs,i)*(-1)**k
bprime[j] = num.real(val)
for j in range(Dp+1):
val = 0.0
for i in range(D+1):
for k in range(i+1):
for l in range(M-i+1):
if k+l == j:
val += comb(i,k)*comb(M-i,l)*a[D-i]*pow(2*fs,i)*(-1)**k
aprime[j] = num.real(val)
#return aprime, bprime
return normalize(bprime, aprime)
开发者ID:shaunhaskey,项目名称:pyMARS,代码行数:33,代码来源:magnetics_generic_funcs_nik.py
示例13: centroid
def centroid(self):
'''Find the centroid x and y from these coefficients'''
xcen = 0.0
ycen = 0.0
fluxtot = self.total_flux()
for i1 in range(self.n1):
if i1%2==0: continue #consider odd i1
for i2 in range(self.n2):
if i2%2!=0: continue # consider even i2
xcen = xcen + np.power(i1+1,0.5)*np.power(2,0.5*(2-i1-i2))* np.power(comb(i1+1,(i1+1)/2)*comb(i2,i2/2),0.5)*self.coeff[i1,i2]
for i1 in range(self.n1):
if i1%2!=0: continue #consider even i1
for i2 in range(self.n2):
if i2%2==0:continue # consider odd i2
ycen = ycen + np.power(i2+1,0.5)*np.power(2,0.5*(2-i2-i1))* np.power(comb(i2+1,(i2+1)/2)*comb(i1,i1/2),0.5)*self.coeff[i1,i2]
xcen = xcen*np.sqrt(pi)*self.beta*self.beta/fluxtot
ycen = ycen*np.sqrt(pi)*self.beta*self.beta/fluxtot
return xcen,ycen
开发者ID:dh4gan,项目名称:shapelet_decomposition,代码行数:29,代码来源:coefficients.py
示例14: multiple_alignment
def multiple_alignment(word_list):
'''Returns the multiple alignment of a given list of words.'''
from itertools import product
from operator import add, mul
from scipy.misc import comb
# There are some issues scoring the first symbols, so force a match here and remove it from the alignment later.
word_list = ['$'+word for word in word_list]
# Initialize scoring and backtrack dictionaries, along with the indices and base score.
S, backtrack = {}, {}
perm_list = list(product([0, -1], repeat=len(word_list)))[1:]
base_score = -1*comb(len(word_list), 2, exact=True)
for index in product(*map(xrange,map(lambda s: len(s) + 1, word_list))):
# We forced a match with the first symbols, so the zero-shell should lead to the zero index.
if reduce(mul, index) == 0:
# Since we forced a match with the first symbol, we want to force starting point to be the zero index.
if sum(index) == 0:
# All symbols match.
S[index] = 0
else:
# Make it smaller than the lowest possible score.
S[index] = 2*base_score*reduce(add, map(len, word_list))
else:
# Use previous scores to determine the best score for the current index.
previous_scores = [S[tuple(map(add, index, perm))] for perm in perm_list]
current_index_scores = []
for perm in perm_list:
chars = [word_list[i][index[i]-1] if perm_value == -1 else '-' for i, perm_value in enumerate(perm)]
current_index_scores.append(base_score + sum([comb(chars.count(ch), 2, exact=True) for ch in set(chars)]))
scores = map(add, previous_scores, current_index_scores)
backtrack[index], S[index] = max(enumerate(scores), key=lambda p: p[1])
# Initialize the alignment and indicies.
alignment = word_list
current_index = map(len, word_list)
# Get the max score.
# Note: The forced match at start of each word does not change the max score, as matched symbols have a score of zero.
max_score = S[tuple(current_index)]
# Quick lambda function to insert indels.
insert_indel = lambda word, i: word[:i] + '-' + word[i:]
# Insert indels to get the alignment.
while reduce(mul, current_index) != 0:
for i, perm_value in enumerate(perm_list[backtrack[tuple(current_index)]]):
if perm_value == 0:
alignment[i] = insert_indel(alignment[i], current_index[i])
else:
current_index[i] -= 1
# Note: We don't need to prepend any indels because we forced a match at the start of all words.
# Remove the forced match from all alignments to recover the correct alignment.
return [str(max_score)] + [str(aligned[1:].seq) for aligned in alignment]
开发者ID:crf1111,项目名称:Bio-Informatics-Learning,代码行数:59,代码来源:Multiple_Alignment.py
示例15: elm_comb
def elm_comb(rho, N, m):
offered_traffic = rho / (1 + rho)
numerator = comb(N-1, m, exact=True)*pow(offered_traffic, m)*pow(1-offered_traffic, N-m)
denominator = 0
for i in range(0, m+1):
denominator += comb(N-1, i, exact=True)*pow(offered_traffic, i)*pow(1-offered_traffic, N-i)
return numerator / denominator
开发者ID:tommyjcarpenter,项目名称:queueing-models,代码行数:8,代码来源:engset_loss_model.py
示例16: rectangles_numbers
def rectangles_numbers(M, N):
'''
正方形是屬於長方形的一種。或者說,正方形是長方形的特例。矩形的定義是四個角皆為 90° 的四邊形。
如果要算不包含正方形的長方形,只需把所有的算出,然後減去正方形的個數。
'''
horizontal = sm.comb(M+1, 2, exact=True)
vertical = sm.comb(N+1, 2, exact=True)
return horizontal * vertical
开发者ID:navoval,项目名称:python_practice_codes,代码行数:8,代码来源:interview_questions.py
示例17: hyperg
def hyperg(k, N, m, n, verbose=False):
"""
k = intersection; m = population 1; n = population 2; N = total population.
"""
exact=1
if verbose and k%100==0: printnow("k: %s" % k)
#return comb(m,k,exact) * comb(N-m,n-k,exact)/ comb(N,n,exact)
return comb(m,k) * comb(N-m,n-k)/ comb(N,n)
开发者ID:marcottelab,项目名称:infer_complexes,代码行数:8,代码来源:utils.py
示例18: runs_prob_odd
def runs_prob_odd(self, r):
n0, n1 = self.n0, self.n1
k = (r+1)//2
tmp0 = comb(n0-1, k-1)
tmp1 = comb(n1-1, k-2)
tmp3 = comb(n0-1, k-2)
tmp4 = comb(n1-1, k-1)
return (tmp0 * tmp1 + tmp3 * tmp4) / self.comball
开发者ID:0ceangypsy,项目名称:statsmodels,代码行数:8,代码来源:runs.py
示例19: hypergeometricown
def hypergeometricown(self, N, K, n, k):
"""
N= total number of genes in population
K= number of GOA
n= select a sample (top 50, bottom half, etc.)
k= number of successes in the sample
"""
return comb(K, k) * comb(N-K, n-k) / comb(N, n)
开发者ID:wimverleyen,项目名称:SAPLING,代码行数:8,代码来源:enrichment.py
示例20: mv_hypergeometric
def mv_hypergeometric(x, m):
"""
x : number of draws for each category.
m : size of each category.
"""
x = np.asarray(x)
m = np.asarray(m)
return log(comb(m, x).prod() / comb(m.sum(), x.sum()))
开发者ID:CamDavidsonPilon,项目名称:pymc,代码行数:8,代码来源:test_distributions.py
注:本文中的scipy.misc.comb函数示例由纯净天空整理自Github/MSDocs等源码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。 |
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