Please help me understand output from discretize

I am trying to understand what discretize is doing, I have data ranging from 0 to 1 and want to bin it in bins of size .01.

X=[.01:.01:1] ; % - example data uniformly going form .01 to 1 in steps of .01  
binEdges=[0:0.01:1]; %- the bind edges
binLocs=discretize(X,binEdges,'IncludedEdge','right'); % 
%I would expect each bin gets 1 value so that bon locs is  [1 2 3 4 5 6 7 8 9 10  etc... up to 100] 

but what I get is

binLocs=

1 2 3 4 5 7 7 8 9 10 11 12 13 14 16 16 17 19 19 20 22 22 23 25 25 26 27 28 30 30 31 32 33 34 35 37 37 38 39 40 41 43 43 44 45 46 47 49 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

why is is repeating some (for example the values .06 and .07 go in bin 7 and nothing in bin 6 ?

I'd like the first bin to b 0<=x<=.01 bin 2 .01<x<=.02 bin 3 .02<x<=.03 etc..

can someone explain why I am getting 2 values in some bins ( i.e. bin 7 and bin 22)

Thanks,

-Jeff

Answer 1

[Torsten]

X=[.01:.01:1] ; % - example data uniformly going form .01 to 1 in steps of .01  
binEdges=[0:0.01:1]; %- the bind edges
X(6)>binEdges(7) & X(6)<=binEdges(8)

ans = logical
   1

X(7)>binEdges(7) & X(7)<=binEdges(8)

ans = logical
   1

Since your data are on the edges of the bins, it's a precision problem in several cases.

[Jeff Spector]

I thought that 'includge edge, right" would take of that. I still don't understand the following :

X(6)

ans =

0.0600

binEdges(7)

ans =

0.0600

.0600>06000

ans =

logical

0

%- this makes sense since they are equal but then

.0600>binEdges(7)

ans =

logical

0

%- ok this is same as above, now replace .0600 with X(6)

X(6)

ans =

0.0600

>> X(6)>binEdges(7)

ans =

logical

1

%- I guess I don't understand why this is happening? Can someone enlighten me. It seems to get it correct if I'm not using "X(6) but rather the value of X(6). This seems insane to me so please help me understand what I am missing.

Thanks,

-Jeff

[Torsten]

X(6) is not 0.0600. It's only printed as output with this limited length.

X=[.01:.01:1] ; % - example data uniformly going form .01 to 1 in steps of .01  
binEdges=[0:0.01:1]; %- the bind edges
binEdges(7)-X(6)

ans = -6.9389e-18

X(6)-binEdges(8)

ans = -0.0100

binEdges(7)-X(7)

ans = -0.0100

X(7)-binEdges(8)

ans = -1.3878e-17

Answer 2

[Walter Roberson]

X=[.01:.01:1] ; % - example data uniformly going form .01 to 1 in steps of .01  
binEdges=[0:0.01:1]; %- the bind edges

X - round(X * 100)/100

ans = 1x100

1.0e-15 * 0 0 0 0 0 0.0069 -0.0139 0 0 -0.0139 0 0 0 0 0.0278 0 0 0.0278 0 0 0.0278 0 0 0.0278 0 0 0 0 0.0555 0

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

binEdges - round(binEdges*100)/100

ans = 1x101

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

So, starting from 0.01 and incrementing by 0.01 does not lead to edges that are exactly what you would naively predict.

[Jeff Spector]

Thanks. I sitll don't really understand where the number are coming from ( i.e. why aren't they .01,.02 etc.. but just using round (X,5) seems to fix the issue, and I can easily tolerate that level of roundig so thanks so much!

-Jeff

[Walter Roberson]

X = 0.01:0.01:1;
Y = (1:100)/100;
X(6:7) - Y(6:7)

ans = 1x2

1.0e-16 * 0.0694 -0.1388

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

fprintf('X(6:7) = %08x %08x\n', typecast(X(6:7), 'uint64'));

X(6:7) = 3faeb851eb851eb9 3fb1eb851eb851eb

fprintf('Y(6:7) = %08x %08x\n', typecast(Y(6:7), 'uint64'));

Y(6:7) = 3faeb851eb851eb8 3fb1eb851eb851ec

fprintf('X5+ = %08x\n', typecast(X(5) + 0.01, 'uint64'))

X5+ = 3faeb851eb851eb9

I would guess that 0:0.01:1 is being treated internally as (0:100)/100 whereas 0.01:0.01:1 is treated as repeated addition, and repeated addition suffers from round-off error.