def longest_ones(x): """ return the indicies of the longest stretch of contiguous ones in x, assuming x is a vector of zeros and ones. If there are two equally long stretches, pick the first """ x = asarray(x) if len(x)==0: return array([]) #print 'x', x ind = find(x==0) if len(ind)==0: return arange(len(x)) if len(ind)==len(x): return array([]) y = zeros( (len(x)+2,), Int) y[1:-1] = x d = diff(y) #print 'd', d up = find(d == 1); dn = find(d == -1); #print 'dn', dn, 'up', up, ind = find( dn-up == max(dn - up)) # pick the first if iterable(ind): ind = ind[0] ind = arange(up[ind], dn[ind]) return ind
def hist(y, bins=10, normed=0): """ Return the histogram of y with bins equally sized bins. If bins is an array, use the bins. Return value is (n,x) where n is the count for each bin in x If normed is False, return the counts in the first element of the return tuple. If normed is True, return the probability density n/(len(y)*dbin) If y has rank>1, it will be raveled Credits: the Numeric 22 documentation """ y = asarray(y) if len(y.shape)>1: y = ravel(y) if not iterable(bins): ymin, ymax = min(y), max(y) if ymin==ymax: ymin -= 0.5 ymax += 0.5 bins = linspace(ymin, ymax, bins) n = searchsorted(sort(y), bins) n = diff(concatenate([n, [len(y)]])) if normed: db = bins[1]-bins[0] return 1/(len(y)*db)*n, bins else: return n, bins
def longest_ones(x): """ return the indicies of the longest stretch of contiguous ones in x, assuming x is a vector of zeros and ones. If there are two equally long stretches, pick the first """ x = asarray(x) if len(x) == 0: return array([]) #print 'x', x ind = find(x == 0) if len(ind) == 0: return arange(len(x)) if len(ind) == len(x): return array([]) y = zeros((len(x) + 2, ), Int) y[1:-1] = x d = diff(y) #print 'd', d up = find(d == 1) dn = find(d == -1) #print 'dn', dn, 'up', up, ind = find(dn - up == max(dn - up)) # pick the first if iterable(ind): ind = ind[0] ind = arange(up[ind], dn[ind]) return ind
def hist(y, bins=10, normed=0): """ Return the histogram of y with bins equally sized bins. If bins is an array, use the bins. Return value is (n,x) where n is the count for each bin in x If normed is False, return the counts in the first element of the return tuple. If normed is True, return the probability density n/(len(y)*dbin) If y has rank>1, it will be raveled Credits: the Numeric 22 documentation """ y = asarray(y) if len(y.shape) > 1: y = ravel(y) if not iterable(bins): ymin, ymax = min(y), max(y) if ymin == ymax: ymin -= 0.5 ymax += 0.5 bins = linspace(ymin, ymax, bins) n = searchsorted(sort(y), bins) n = diff(concatenate([n, [len(y)]])) if normed: db = bins[1] - bins[0] return 1 / (len(y) * db) * n, bins else: return n, bins
def longest_contiguous_ones(x): """ return the indicies of the longest stretch of contiguous ones in x, assuming x is a vector of zeros and ones. """ if len(x)==0: return array([]) ind = find(x==0) if len(ind)==0: return arange(len(x)) if len(ind)==len(x): return array([]) y = zeros( (len(x)+2,), x.typecode()) y[1:-1] = x dif = diff(y) up = find(dif == 1); dn = find(dif == -1); ind = find( dn-up == max(dn - up)) ind = arange(take(up, ind), take(dn, ind)) return ind
def longest_contiguous_ones(x): """ return the indicies of the longest stretch of contiguous ones in x, assuming x is a vector of zeros and ones. """ if len(x) == 0: return array([]) ind = find(x == 0) if len(ind) == 0: return arange(len(x)) if len(ind) == len(x): return array([]) y = zeros((len(x) + 2, ), typecode(x)) y[1:-1] = x dif = diff(y) up = find(dif == 1) dn = find(dif == -1) ind = find(dn - up == max(dn - up)) ind = arange(take(up, ind), take(dn, ind)) return ind
def trapz(x, y): if len(x)!=len(y): raise ValueError, 'x and y must have the same length' if len(x)<2: raise ValueError, 'x and y must have > 1 element' return asum(0.5*diff(x)*(y[1:]+y[:-1]))
def trapz(x, y): if len(x) != len(y): raise ValueError, 'x and y must have the same length' if len(x) < 2: raise ValueError, 'x and y must have > 1 element' return asum(0.5 * diff(x) * (y[1:] + y[:-1]))