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copied!<p>This is not exactly the answer you were looking for, but this is my 50c answer...</p> <pre> A = {1:'a1',2:'a2',3:'a3',4:'a4'} B = {0.7:'b1',0.9:'b2',1.7:'b3',2:'b4', 2.4:'b5'} C = {} # result # find altitude in A that is the closest to altitude in B def findAltitude( altB,A): toto = [ ((alt-altB)**2,alt) for alt in A.keys() ] toto.sort() return toto[0][1] #iter on each altitude of B for altB,valueB in B.iteritems(): altC = findAltitude( altB,A) if altC in C: C[altC].append(valueB) else: C[altC] = [valueB,] # then do the median operation #for altC,valueC in C.iteritems(): # C[altC] = map( median, valueC ) # where median is your median function print C </pre> <p>It is NOT the best solution at all (specially if you have a lot of values), but only the fastest to write...</p> <p>In fact, it depends how your datas are stored. Dictionnary is not the best choice.</p> <p>It is more interesting/clever to use the fact that your altitudes are sorted. You should provide more details on how your datas are stored (array with numpy ?) </p> <p>==== Edit ====</p> <p>I still do not know how your datas are, but let's try something more "clever", based on the fact that your altitudes are sorted.</p> <pre><code>from numpy import * from pylab import * from scipy.stats import nanmedian # add val into C at the end of C or in the last place (depending if alt already exists in C or not) def addto(C,val,alt): if C and C[-1][0]==alt: C[-1][1].append(valB) else: C.append( (alt,[valB,] )) # values A_Alt = array([1,2,3,4]) A_Val = array([.33, .5, .6, .8]) B_Alt = array([.7, 0.9, 1.7, 2., 2.4, 2.9, 3.1, 3.2, 3.9, 4.1]) B_Val = array([.3, NaN, .8, .6, .7, .4, .3, NaN, .99, 1.3]) #intermediate list of tuple (altitude, list_of_values) C= [] #iterator on A Aa = iter(A_Alt) ainf = Aa.next() asup = Aa.next() # two first values of A_Alt #iterator on B Ba = iter(B_Alt) Bv = iter(B_Val) # regrid try: while True: altB = Ba.next() valB = Bv.next() # find ainf and asup in A_Alt such that ainf < altB < asup while asup<altB: try: ainf,asup = asup, Aa.next() except StopIteration: break # find closest if abs(ainf-altB)<=abs(asup-altB): addto(C, valB, ainf) else: addto(C, valB, asup) except StopIteration: pass # do the median res = [ nanmedian(k[1]) for k in C ] print res </code></pre> <p>The idea is then to iterate over the two vectors/lists of altitudes, and for each altitude of B, find the two altitudes of A that surround it. Then, it is easy to find the closest...</p> <p>This is less readable than Daan's solution, but it should be more efficient (linear in the size of your datas).</p> <p>You just need to modify if your datas are not stored like that.</p>

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