nan

While writing some test cases, and some of the tests check for the result of a NaN. I tried using std::isnan but the assert failes: Assertion `std::isnan(x)' failed. After printing the value of x , it turned out it's negative NaN ( -nan ) which is totally acceptable in my case. After trying to use the fact that NaN != NaN and using assert(x == x) , the compiler does me a 'favor' and optimises the assert away. Making my own isNaN function is being optimised away as well. How can I check for both equality of NaN and -NaN? This is embarrassing. The reason the compiler (GCC in this case) was

I was asked to filter out NaN, null, 0, false in an array . Luckily I have answered the question. function bouncer(arr) { function filterer(arr) { return arr > 0|| isNaN(arr) === true; } arr = arr.filter(filterer); return arr; } //example input bouncer([0, 1, 2, 3, 'ate', '', false]); //output [1, 2, 3, 'ate'] but the thing is I really don't know how I came up with the answer or rather I don't know how it works. Specially on arr > 0 how did the filter know that arr is alread on arr[1], arr[2], etc.. without using a loop to iterate in all array. or can simply just explain on how to code works.

I have a pandas dataframe (df), and I want to do something like: newdf = df[(df.var1 == 'a') & (df.var2 == NaN)] I've tried replacing NaN with np.NaN , or 'NaN' or 'nan' etc, but nothing evaluates to True. There's no pd.NaN . I can use df.fillna(np.nan) before evaluating the above expression but that feels hackish and I wonder if it will interfere with other pandas operations that rely on being able to identify pandas-format NaN's later. I get the feeling there should be an easy answer to this question, but somehow it has eluded me. Any advice is appreciated. Thank you. Mark Whitfield This

I've got a numpy array filled mostly with real numbers, but there is a few nan values in it as well. How can I replace the nan s with averages of columns where they are? No loops required: print(a) [[ 0.93230948 nan 0.47773439 0.76998063] [ 0.94460779 0.87882456 0.79615838 0.56282885] [ 0.94272934 0.48615268 0.06196785 nan] [ 0.64940216 0.74414127 nan nan]] #Obtain mean of columns as you need, nanmean is just convenient. col_mean = np.nanmean(a, axis=0) print(col_mean) [ 0.86726219 0.7030395 0.44528687 0.66640474] #Find indicies that you need to replace inds = np.where(np.isnan(a)) #Place