accumulate

With using python library numpy, it is possible to use the function cumprod to evaluate cumulative products, e.g. a = np.array([1,2,3,4,2]) np.cumprod(a) gives array([ 1, 2, 6, 24, 48]) It is indeed possible to apply this function only along one axis. I would like to do the same with matrices (represented as numpy arrays), e.g. if I have S0 = np.array([[1, 0], [0, 1]]) Sx = np.array([[0, 1], [1, 0]]) Sy = np.array([[0, -1j], [1j, 0]]) Sz = np.array([[1, 0], [0, -1]]) and b = np.array([S0, Sx, Sy, Sz]) then I would like to have a cumprod -like function which gives np.array([S0, S0.dot(Sx), S0

In an effort to be as lazy as possible I read in a matrix as vector< vector<double> > data ( rows, vector<double> ( columns ) ); and try to use as many STL goodies as I can. One thing I need to do next is to compute the row means. In C-style programming that would be vector<double> rowmeans( data.size() ); for ( int i=0; i<data.size(); i++ ) for ( int j=0; j<data[i].size(); j++ ) rowmeans[i] += data[i][j]/data[i].size(); In In C++, how to compute the mean of a vector of integers using a vector view and gsl_stats_mean? it is explained that for a vector of numbers you can compute a vector mean

I am trying to combine std::accumulate with std::min . Something like this (won't compile): vector<int> V{2,1,3}; cout << accumulate(V.begin()+1, V.end(), V.front(), std::min<int>); Is it possible? Is it possible to do without writing wrapper functor for std::min ? I know that I can do this with lambdas: vector<int> V{2,1,3}; cout << std::accumulate( V.begin()+1, V.end(), V.front(), [](int a,int b){ return min(a,b);} ); And I know there is std::min_element . I am not trying to find min element, I need to combine std::accumulate with std::min (or ::min ) for my library which allows function

I was wondering if there is any pandas equivalent to cumsum() or cummax() etc. for median: e.g. cummedian() . So that if I have, for example this dataframe: a 1 5 2 7 3 6 4 4 what I want is something like: df['a'].cummedian() which should output: 5 6 6 5.5 You can use expanding.median - df.a.expanding().median() 1 5.0 2 6.0 3 6.0 4 5.5 Name: a, dtype: float64 Timings df = pd.DataFrame({'a' : np.arange(1000000)}) %timeit df['a'].apply(cummedian()) 1 loop, best of 3: 1.69 s per loop %timeit df.a.expanding().median() 1 loop, best of 3: 838 ms per loop The winner is expanding.median by a huge