If I type !i.10 it gives the factorial of first 10 numbers.
However if I try to add a column of ordinal numbers >/.!i.10, 1+i.10, then J freezes or I get an "Out of memory" error. How do I create custom tables?
I think that what is happening is that you are creating something much bigger than you expect. Taking it in steps:
1+ i. 10 NB. list of 1 to 10
1 2 3 4 5 6 7 8 9 10
10 , 1+ i. 10 NB. 10 prepended
10 1 2 3 4 5 6 7 8 9 10
i. 10 , 1+ i. 10 NB. creates an 11 dimension array with shape 10 1 2 3 4 5 6 7 8 9 10 and largest value of 36287999
When you apply ! to that i. 10 , 1+ i. 10 you get some very large numbers. I am not sure what you are trying to do with the leading >/.
Is this what you had in mind?
(!1 + i.10),. (1+i.10) NB. using parenthesis to isolate operations
1 1
2 2
6 3
24 4
120 5
720 6
5040 7
40320 8
362880 9
3.6288e6 10
To give extended type and get rid of the 3.6288e6 we can use x:
(x:!1 + i.10),. (1+i.10)
1 1
2 2
6 3
24 4
120 5
720 6
5040 7
40320 8
362880 9
3628800 10
or tacit
(x:#! ,. ]) # (1+i.) 10
1 1
2 2
6 3
24 4
120 5
720 6
5040 7
40320 8
362880 9
3628800 10
Or a version I find a little better
([: (,.~ !) 1x + i.) 10
1 1
2 2
6 3
24 4
120 5
720 6
5040 7
40320 8
362880 9
3628800 10
Related
I need to do a complicated vlookup/maxif type of selection. The data I have is as below
Row Col G Col H Col I colJ col K
1 Bench Strip Block BenchAbove BenchBelow
2 1 1 4
3 1 1 5
4 1 1 6
5 1 1 7
6 1 1 8
7 8 1 4 ?? ??
8 8 1 5
9 8 1 6
10 8 1 7
11 8 1 8
12 9 1 4
13 9 1 5
14 9 1 6
15 9 1 7
.....this list is long ( this is a sample only)
For every combination of (Strip, block) like say (1,4) There are benches like 1, 8 and 9. So bench above for 8 is 1 and bench below for 8 is 9. I need to determine the bench above and bench below for each row. There are no bench above 1 and no bench below 9.
I dont think vlookup is the solution here. Not sure if MAX(IF..) can help either. What would be the best formulae to obtain say on row 7, block combination is 1,4. The bench in question is 8. The bench above is 1 and bench below is 9. So 2 formulae will be required on Col J and Col I above.
The expected answer for the above sample data is :
Row Col G Col H Col I colJ col K
1 Bench Strip Block BenchAbove BenchBelow
2 1 1 4 - 8
3 1 1 5 - 8
4 1 1 6 - 8
5 1 1 7 - 8
6 1 1 8 - 8
7 8 1 4 1 9
8 8 1 5 1 9
9 8 1 6 1 9
10 8 1 7 1 9
11 8 1 8 1 9
12 9 1 4 8 -
13 9 1 5 8 -
14 9 1 6 8 -
15 9 1 7 8 -
Maybe in J2:
=IFERROR(LOOKUP(2,1/((H$1:H1=H2)*(I$1:I1=I2)),G$1:G1),"-")
In K2:
=IFERROR(INDEX(G3:G$16,MATCH(1,INDEX((H3:H$16=H2)*(I3:I$16=I2),),0)),"-")
However, I find your question a bit confusing so this answer might be a bit off.
If you have Office365 then you can use MAXIFS(), MINIFS() easily to get BenchAbove and BenchBelow. Try-
=MAXIFS(A2:A15,B2:B15,B7,C2:C15,C7)
=MINIFS(A2:A15,B2:B15,B7,C2:C15,C7)
EDIT: Solution for Excel-2016
Try below formula-
=INDEX($A$2:$A$15,AGGREGATE(14,6,ROW($A$2:$A$15)-ROW($A$1)/(($B$2:$B$15=B7)*($C$2:$C$15=C7)),ROW(1:1)))
=INDEX($A$2:$A$15,AGGREGATE(15,6,ROW($A$2:$A$15)-ROW($A$1)/(($B$2:$B$15=B7)*($C$2:$C$15=C7)),ROW(1:1)))
I'm trying to do this program where given a number N, one has to print out the decimal, octal, hexadecimal and binary for all the numbers in range 1 to N. The trouble is that the platform requires the solution in a particular format.
Suppose the number is 17, so the output should be like :
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
10 12 A 1010
11 13 B 1011
12 14 C 1100
13 15 D 1101
14 16 E 1110
15 17 F 1111
16 20 10 10000
17 21 11 10001
For 7 it would be like :
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
If you notice, the above is required to be printed in a way that the decimal, octal and hexadecimal numbers need a minimum of 2 spaces at their left whereas the binary numbers need at least one space at their left. Now, as the length of the numbers increase the space needs to be given accordingly such that the minimum space is there even for the max length number. So, how do I print them using a variable space? So far I have tried this :
Code
def print_formatted(number):
space=len(str(bin(number))[2:])
for i in range(1,number+1):
print('{:2d}'.format(i), end='')
print('{:>3s}'.format(str(oct(i))[2:]), end='')
print('{:>3s}'.format(str(hex(i))[2:]), end='')
print('{:>'+str(space)+'s}'.format(str(bin(i))[2:]))
print_formatted(17)
Here, I just tried doing the required with just the binary numbers but it's giving me an error
print('{:>'+str(space)+'s}'.format(str(bin(i))[2:]))
ValueError: Single '}' encountered in format string
Is there any fix/alternative for this?
Your problem is operator order - the + for string concattenation is weaker then the method call in
'{:>' + str(space) + 's}'.format(str(bin(i))[2:])
. Thats why you call the .format(...) only on "s}" - not the whole string. And thats where the
ValueError: Single '}' encountered in format string
comes from.
Putting the complete formatstring into parenthesis before applying .format to it fixes that.
You also need 1 more space for binary and can skip some str() that are not needed:
def print_formatted(number):
space=len(str(bin(number))[2:])+1 # fix here
for i in range(1,number+1):
print('{:2d}'.format(i), end='')
print('{:>3s}'.format(oct(i)[2:]), end='')
print('{:>3s}'.format(hex(i)[2:]), end='')
print(('{:>'+str(space)+'s}').format(bin(i)[2:])) # fix here
print_formatted(17)
Output:
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
10 12 a 1010
11 13 b 1011
12 14 c 1100
13 15 d 1101
14 16 e 1110
15 17 f 1111
16 20 10 10000
17 21 11 10001
From your given output above you might need to prepend this by 2 spaces - not sure if its a formatting error in your output above or part of the restrictions.
You could also shorten this by using f-strings (and removing superflous str() around bin, oct, hex: they all return a strings already).
Then you need to calculate the the numbers you use to your space out your input values:
def print_formatted(number):
de,bi,oc,he = len(str(number)), len(bin(number)), len(oct(number)), len(hex(number))
for i in range(1,number+1):
print(f' {i:{de}d}{oct(i)[2:]:>{oc}s}{hex(i)[2:]:>{he}s}{bin(i)[2:]:>{bi}s}')
print_formatted(26)
to accomodate other values then 17, f.e. 128:
1 1 1 1
2 2 2 10
3 3 3 11
...
8 10 8 1000
...
16 20 10 10000
...
32 40 20 100000
...
64 100 40 1000000
...
128 200 80 10000000
Input is a number, e.g. 9 and I want to print decimal, octal, hex and binary value from 1 to 9 like:
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
How can I achieve this in python3 using syntax like
dm, oc, hx, bn = len(str(9)), len(bin(9)[2:]), ...
print("{:dm%d} {:oc%s}" % (i, oct(i[2:]))
I mean if number is 999 so I want decimal 10 to be printed like ' 10' and binary equivalent of 999 is 1111100111 so I want 10 like ' 1010'.
You can use str.format() and its mini-language to do the whole thing for you:
for i in range(1, 10):
print("{v} {v:>6o} {v:>6x} {v:>6b}".format(v=i))
Which will print:
1 1 1 1
2 2 2 10
3 3 3 11
4 4 4 100
5 5 5 101
6 6 6 110
7 7 7 111
8 10 8 1000
9 11 9 1001
UPDATE: To define field 'widths' in a variable you can use a format-within-format structure:
w = 5 # field width, i.e. offset to the right for all octal/hex/binary values
for i in range(1, 10):
print("{v} {v:>{w}o} {v:>{w}x} {v:>{w}b}".format(v=i, w=w))
Or define a different width variable for each field type if you want them non-uniformly spaced.
Btw. since you've tagged your question with python-3.x, if you're using Python 3.6 or newer, you can use Literal String Interpolation to simplify it even more:
w = 5 # field width, i.e. offset to the right for all octal/hex/binary values
for v in range(1, 10):
print(f"{v} {v:>{w}o} {v:>{w}x} {v:>{w}b}")
Consider a dyadic verb g, defined in terms of a dyadic verb f:
g=. [ f&.|: f
Is it possible to rewrite g so that the f term appears only once, but the behavior is unchanged?
UPDATE: Local Context
This question came up as part of my solution to this problem, which "expanding" a matrix in both directions like so:
Original Matrix
1 2 3
4 5 6
7 8 9
Expanded Matrix
1 1 1 1 2 3 3 3 3
1 1 1 1 2 3 3 3 3
1 1 1 1 2 3 3 3 3
1 1 1 1 2 3 3 3 3
4 4 4 4 5 6 6 6 6
7 7 7 7 8 9 9 9 9
7 7 7 7 8 9 9 9 9
7 7 7 7 8 9 9 9 9
7 7 7 7 8 9 9 9 9
My solution was to expand the matrix rows first using:
f=. ([ # ,:#{.#]) , ] , [ # ,:#{:#]
And then to apply that same solution under the transpose to expand the columns of the already row-expanded matrix:
3 ([ f&.|: f) m
And I noticed that it wasn't possible to write my solution with making the temporary verb f, or repeating its definition inline...
Try it online!
Knowing the context helps. You can also approach this using (|:#f)^:(+: x) y. A tacit (and golfed) solution would be 0&(|:{.,],{:)~+:.
(>: i. 3 3) (0&(|:{.,],{:)~+:) 2
1 1 1 2 3 3 3
1 1 1 2 3 3 3
1 1 1 2 3 3 3
4 4 4 5 6 6 6
7 7 7 8 9 9 9
7 7 7 8 9 9 9
7 7 7 8 9 9 9
I don't think it is possible. The right tine is going to be the result of x f y and the left tine is x The middle tine will transpose and apply f to the arguments and then transpose the result back. If you take the right f out then there is not a way to have x f y and if the middle f is removed then you do not have f applied to the transpose.
My guess is that you are looking for a primitive that will accomplish the same result with only one mention of f, but I don't know of one.
Knowing the J community someone will prove me wrong!
I have 15 datafiles with unequal row sizes, but number of columns in each file is same. e.g.
ifile1.dat ifile2.dat ifile3.dat and so on ............
0 0 0 0 1 6
1 2 5 3 2 7
2 5 6 10 4 6
5 2 8 9 5 9
10 2 10 3 8 2
In each file 1st column represents the index number.
I would like to compute average of all these files for each index number in column 1. i.e.
ofile.txt
0 0 [This is computed as (0+0)/2]
1 4 [This is computed as (2+6)/2]
2 6 [This is computed as (5+7)/2]
3 [no value]
4 6 [This is computed as (6)/1]
5 4.66 [This is computed as (2+3+9)/3]
6 10
7
8 5.5
9
10 2.5
I can't think of any simple method to do it. I was thinking of a method, but seems very lengthy. Taking the average after converting all the files with same row size, .e.g.
ifile1.dat ifile2.dat ifile3.dat and so on ............
0 0 0 0 0 0
1 2 1 1 6
2 5 2 2 7
3 3 3
4 4 4 6
5 2 5 3 5 9
6 6 10 6
7 7 7
8 8 9 8 2
9 9 9
10 2 10 3 10
$ awk '{s[$1]+=$2; c[$1]++;} END{for (i in s) print i,s[i]/c[i];}' ifile*.dat
0 0
1 4
2 6
4 6
5 4.66667
6 10
8 5.5
10 2.5
In the above code, there are two arrays, s and c. s[i] is the sum of all entries with index i and c[i] is the number of entries with index i. After we have read all the files, we print the average, s[i]/c[i], for each index i.