Compute percentages in a list

Question

I have a data file which contains thousands of lines and each line has eight elements. Here is a small sample of the data file

-4.00      -0.80   0.1886024468848907E+01   0.1467147621657460E+01   -.1217067274319363E+01   0.7206100000000000E+03   0.7693457688734395E-12    5
-4.00      -0.70   0.1430357986632780E+01   -.1404093461650013E+01   -.1742223680347601E+01   0.1824700000000000E+03   0.8439003681169850E-12    8
-4.00      -0.60   -.1324719768465547E+01   0.1740130076002850E+01   0.1497978622206873E+01   0.5479900000000000E+03   0.4264485578634903E-14    2
-4.00      -0.50   0.1358536876189560E+01   -.1580696533502541E+01   0.1621539980382560E+01   0.2881100000000000E+03   0.1319885603098060E-13    4
-4.00      -0.40   -.1588487538231399E+01   0.1275577589608218E+01   0.1707607247015512E+01   0.1337500000000000E+03   0.1057878487713421E-12    2
-4.00      -0.30   0.1755414125374284E+01   0.1332710827520201E+01   0.1477984475201826E+01   0.6426400000000000E+03   0.1459764022611444E-13    1
-4.00      -0.20   0.1245972697710741E+01   0.1540633564543885E+01   0.1777167629372046E+01   0.6112200000000000E+03   0.5718386586661092E-13    1
-4.00      -0.10   -.1311461418732105E+01   -.1594149065989313E+01   0.1661344176980193E+01   0.3507800000000000E+03   0.6765588799377521E-14    3

I read this file using

data = ReadList["data.out", Number, RecordLists -> True];

The total length of the list is obtained, of course as

ntot = Length[data];

The list contains eight elements per row and the last of them is an integer taking values in the interval [0,8]. What I want is the following:

(a). Count how many rows have 0 value at the last element (let's suppose there are n0), how many have 1, 2, 3, ... , 8. Then calculate the corresponding percentages per0 = n0/ntot, per1 = n1/ntot, etc. It could be nice if this was inside a DO loop with i = 0,8.

(b). Count again percentages but using more than one criteria this time. For example, count how many rows have 1 at the last element and the value of the seventh element is smaller than 10^{-4}.

Any suggestions?

EDIT

Using's @Kuba's solution we have

{{5, 656}, {8, 640}, {2, 673}, {4, 663}, {1, 673}, {3, 663}, {6, 656}, {7, 640}, {0, 19}}

Is it possible to divide each sum automatically with ntot thus obtaining the percentages?

{{5, 656/ntot}, {8, 640/ntot}, {2, 673/ntot}, {4, 663/ntot}, {1, 673/ntot}, {3, 663/ntot}, {6, 656/ntot}, {7,640/ntot}, {0, 19/ntot}}

Also it would be great if they were sorted from 0 to 8 not randomly as they are now.

Answer 1

There are many ways to acheive this:

SetDirectory@NotebookDirectory[];
data = Import["list.txt", "Table"] (*I saved your sample in txt file*)

ntot = Length@data;

temp = SortBy[Tally@data[[;; , -1]], 1];
temp[[;; , 2]] = temp[[;; , 2]] 100/ntot // N; 
(*I've multiplied by 100 to get % value not the ratio*)
temp

{{1, 25.}, {2, 25.}, {3, 12.5}, {4, 12.5}, {5, 12.5}, {8, 12.5}}

What can be useful, but the difference is it will "count" also what is not there (e.g 7):

viahist = HistogramList[data[[;; , 8]], {1}, "Probability"]

{{0, 1, 2, 3, 4, 5, 6, 7, 8, 9},
  {0., 0.25, 0.25, 0.125, 0.125, 0.125, 0., 0., 0.125}}

notice that first part is a list of intervals limits so at the end:

{Most@#, #2 100} & @@ viahist // Transpose

{{0, 0.}, {1, 25.}, {2, 25.}, {3, 12.5}, {4, 12.5}, {5, 12.5}, 
{6, 0.}, {7, 0.}, {8, 12.5}}

---

For your second question, one way to do this is:

Length@Select[data, #[[7]] < 10^(-13) && #[[8]] == 2 &] 100 / ntot // N

12.5

---

Answer 2

You could write functions to generalize this but for your specific problem:

data = Import["list.txt", "Data"];
ntot = Length[data];

(a) to get percentages of elements in the last column (you could optionally Sort this list too)

Inner[Times, Tally[data[[All, -1]]], {1, 100./ntot}, List]

Note that I used Inner here but normally I use Part for this sort of thing but Kuba did something similar:

tmp = Tally[data[[All, -1]]];
tmp[[All, 2]] = tmp[[All, 2]]*100./ntot;

(b) to get percentage when last column is 1 and a condition is applied on the second last column

Length[Cases[data, {__, z_ /; z < 10^-4, 1}]]*100./ntot