Anyone got any idea why chibi-scheme throws the error below while attempting to use the
chibi-loop up-from function:
ERROR: car: not a pair: #<opcode cons>
ERROR in set-dk! on line 590 of file /usr/local/share/chibi/init-7.scm: cdr: not a pair: ()
I am using the syntax in the following way:
(import (chibi loop))
>(up-from 0 (to 20) (by 2))
You can't use up-from standalone like that. It only works inside of a loop form:
> (loop ((for x (up-from 0 (to 20) (by 2))))
(display x)
(newline))
0
2
4
6
8
10
12
14
16
18
Related
My issue is, i don't know how to use the output of a function properly. The output contains multiple lines (j = column , i = testresult)
I want to use the output for some other rules in other functions. (eg. if (i) testresult > 5 then something)
I have a function with two loops. The function goes threw every column and test something. This works fine.
def test():
scope = range(10)
scope2 = range(len(df1.columns))
for (j) in scope2:
for (i) in scope:
if df1.iloc[:,[j]].shift(i).loc[selected_week].item() > df1.iloc[:,[j]].shift(i+1).loc[selected_week].item():
i + 1
else:
print(j,i)
break
Output:
test()
1 0
2 3
3 3
4 1
5 0
6 6
7 0
8 1
9 0
10 1
11 1
12 0
13 0
14 0
15 0
I tried to convert it to list, dataframe etc. However, i miss something here.
What is the best way for that?
Thank you!
A fix of your code would be:
def test():
out = []
scope = range(10)
scope2 = range(len(df1.columns))
for j in scope2:
for i in scope:
if df1.iloc[:,[j]].shift(i).loc[selected_week].item() <= df1.iloc[:,[j]].shift(i+1).loc[selected_week].item():
out.append([i, j])
return pd.DataFrame(out)
out = test()
But you probably don't want to use loops as it's slow, please clarify what is your input with a minimal reproducible example and what you are trying to achieve (expected output and logic), we can probably make it a vectorized solution.
Suppose I define a lazy, infinite array using a triangular reduction at the REPL, with a single element pasted onto the front:
> my #s = 0, |[\+] (1, 2 ... *)
[...]
I can print out the first few elements:
> #s[^10]
(0 1 3 6 10 15 21 28 36 45)
I'd like to move the zero element inside the reduction like so:
> my #s = [\+] (0, |(1, 2 ... *))
However, in response to this, the REPL hangs, presumably by trying to evaluate the infinite list.
If I do it in separate steps, it works:
> my #s = 0, |(1, 2 ... *)
[...]
> ([\+] #s)[^10]
(0 1 3 6 10 15 21 28 36 45)
Why doesn't the way that doesn't work...work?
Short answer:
It is probably a bug.
Long answer:
(1, 2 ... *) produces a lazy sequence because it is obviously infinite, but somehow that is not making the resulting sequence from being marked as lazy.
Putting a sequence into an array #s causes it to be eagerly evaluated unless it is marked as being lazy.
Quick fix:
Append lazy to the front.
> my #s = [\+] lazy 0, |(1, 2 ... *)
[...]
> #s[^10]
(0 1 3 6 10 15 21 28 36 45)
How can I export data from Excel to GAMS? I have a set i and parameter b(i).
b(0)= 30,...,b(10)=18 are:
0 30
1 17
2 21
3 32
4 19
5 29
6 24
7 20
8 23
9 27
10 18
I have an Excel file with name "Book1", my code results in ERROR 409:
Unrecognizable item - skip to find a new statement
looking for a ';' or a key word to get started again
Why? What can I do? This is my code:
set i/1*10/
parameter b(i)
$call =xls2gms r=sheet2!B3:C13 i=Book1.xlsx o=set.inc
$include set.inc
;
You need to add semicolons after each command. The only exception are the $ commands compile time commands. Fx:
set i /1*10/;
parameter b(i);
$call =xls2gms r=sheet2!B3:C13 i=Book1.xlsx o=set.inc
$include set.inc
my #numbers = <4 8 15 16 23 42>;
this works:
.say for #numbers[0..2]
# 4
# 8
# 15
but this doesn't:
my $range = 0..2;
.say for #numbers[$range];
# 16
the subscript seems to be interpreting $range as the number of elements in the range (3). what gives?
Working as intended. Flatten the range object into a list with #numbers[|$range] or use binding on Range objects to hand them around. https://docs.perl6.org will be updated shortly.
On Fri Jul 22 15:34:02 2016, gfldex wrote:
> my #numbers = <4 8 15 16 23 42>; my $range = 0..2; .say for
> #numbers[$range];
> # OUTPUT«16»
> # expected:
> # OUTPUT«4815»
>
This is correct, and part of the "Scalar container implies item" rule.
Changing it would break things like the second evaluation here:
> my #x = 1..10; my #y := 1..3; #x[#y]
(2 3 4)
> #x[item #y]
4
Noting that since a range can bind to #y in a signature, then Range being a
special case would make an expression like #x[$(#arr-param)]
unpredictable in its semantics.
> # also binding to $range provides the expected result
> my #numbers = <4 8 15 16 23 42>; my $range := 0..2; .say for
> #numbers[$range];
> # OUTPUT«4815»
> y
This is also expected, since with binding there is no Scalar container to
enforce treatment as an item.
So, all here is working as designed.
A symbol bound to a Scalar container yields one thing
Options for getting what you want include:
Prefix with # to get a plural view of the single thing: numbers[#$range]; OR
declare the range variable differently so it works directly
For the latter option, consider the following:
# Bind the symbol `numbers` to the value 1..10:
my \numbers = [0,1,2,3,4,5,6,7,8,9,10];
# Bind the symbol `rangeA` to the value 1..10:
my \rangeA := 1..10;
# Bind the symbol `rangeB` to the value 1..10:
my \rangeB = 1..10;
# Bind the symbol `$rangeC` to the value 1..10:
my $rangeC := 1..10;
# Bind the symbol `$rangeD` to a Scalar container
# and then store the value 1..10 in it:`
my $rangeD = 1..10;
# Bind the symbol `#rangeE` to the value 1..10:
my #rangeE := 1..10;
# Bind the symbol `#rangeF` to an Array container and then
# store 1 thru 10 in the Scalar containers 1 thru 10 inside the Array
my #rangeF = 1..10;
say numbers[rangeA]; # (1 2 3 4 5 6 7 8 9 10)
say numbers[rangeB]; # (1 2 3 4 5 6 7 8 9 10)
say numbers[$rangeC]; # (1 2 3 4 5 6 7 8 9 10)
say numbers[$rangeD]; # 10
say numbers[#rangeE]; # (1 2 3 4 5 6 7 8 9 10)
say numbers[#rangeF]; # (1 2 3 4 5 6 7 8 9 10)
A symbol that's bound to a Scalar container ($rangeD) always yields a single value. In a [...] subscript that single value must be a number. And a range, treated as a single number, yields the length of that range.
I have the following code in VBA:
For n = 1 To 15
Cells(n, 8) = Application.Combin(2 * n, n)
next n
I want the n in the cells(n,8) to have an incerement 2, so the code skips a line after each entry.
Is it possible to have an other increment variable in this same loop that jumps 2 at once?
Thanks in advance!
EDIT: after reading the comment: I think what is needed is a counter to count, 1,2,3,4,5,6...15, and another one to count 1,3,5,7...15
For that, here is what is need to be done:
basically, you want the first iterator to be a normal counter,
and the second iterator to be odd numbers only.
So here is a simply input output table
input output
----- -----
1 1
2 3
3 5
4 7
5 9
6 11
From the above, we can deduce the formula needed to convert the input into the desired output: output = (input x 2) -1
And so, we can re-write our for loop to be like so:
For n=1 to 15
Cells(n,8) = Application.Combin(2*n-1,n)
Next
============= End of Edit =========================
Simply, use the keyword STEP in the for loop
For n = 1 To 15 STEP 2 'STEP can also be negative
'but you have to reverse the starting, and endin
'value of the iterator
The values for n will be: 1, 3, 5, 7, 9, 11 , 13, 15
Alternatively, use a local variable inside the for loop for that purpose (in-case you want the loop to execute 15 times)
For n=1 to 15
i = n + 1
Cells(i,8) = Application.Combine(2*n,n)
Next