Re: On partitioning lists by intervals

*To*: mathgroup at smc.vnet.net*Subject*: [mg80447] Re: On partitioning lists by intervals*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Wed, 22 Aug 2007 04:55:05 -0400 (EDT)*Organization*: The Open University, Milton Keynes, UK*References*: <faea5n$ejf$1@smc.vnet.net>

Mauricio Esteban Cuak wrote: <snip> > I have a list of say 500 different random values. I need to divide it > in n intervals of fixed length (for example, the lowest value is 0 and > the maximum 100 so I need to get sublists of values that go from 0 to > 10, 10 to 20, etc.) One way to do that is with Select and Table, testing interval membership and sorting (if needed) the values. For instance, In[1]:= data = RandomReal[{0, 100}, {50}]; data = Sort@data; Table[Select[data, IntervalMemberQ[Interval[{m, m + 10}], #] &], {m, 0, 90, 10}] Out[3]= {{2.64545, 2.88612, 6.95366, 8.94619}, {}, {21.2125, 22.5317, 23.3286, 23.5521, 27.0041, 27.0368}, {31.8472, 34.0783, 35.9797, 38.0444}, {40.1175, 40.9223, 44.6731, 46.3196, 47.9082, 48.3849, 49.7482}, {50.6901, 51.3209, 52.0433, 53.6529, 57.0734, 59.8382}, {60.2186, 62.8556, 64.1785, 66.705, 68.0119, 68.1914}, {70.2759, 73.2949, 75.0208, 75.026, 75.269, 76.8917, 77.0392}, {80.4942, 85.8137, 87.242, 87.5028, 88.4888, 89.8753}, {91.4804, 93.1612, 98.4789, 99.0704}} -- Jean-Marc