Can someone explain, why I am not getting expected results?
awk '
# I know there is delete array, but this is more portable
# that is what docs are saying, anyway wanted to test it out.
function delarray(a, i)
{
for (i in a)
delete a[i]
}
BEGIN {
a[3]=""
a[4]=""
for (e in a)
print e
delarray(a)
for (e in a)
print ".."
print e
}
'
Executing the above script, I expected to see:
3
4
..(nothing here)
I used .. thinking I won't see anything else because of
deleted array values so just to see .. as placeholder)
,but the actual output I see is:
3
4
4 #(why this?, and where are two dots?)
,also exit code was 1, why is that?
Your delete function worked.
Since you're missing the braces around your for (e in a) loop, it only contains the print ".." statement, which is why you don't see any dots.
The print e command simply prints the last value that was assigned to e (from the previous for (e in a) loop), which is 4.
But your function is not very useful since virtually all versions of awk allow the delete a command without an index. It's in the POSIX standard.
You are missing curly braces around your second loop.
for (e in a) {
print ".."
print e
}
Output:
3
4
Related
I'm pretty new to Perl and this is my most complex project yet. Apologies if any parts of my explanation don't make sense or I miss something out - I'll be happy to provide further clarification. It's only one line of code that's causing me an issue.
The Aim:
I have a text file that contains a single column of data. It reads like this:
0
a,a,b,a
b,b,b,a
1
a,b,b,a
b,b,b,a
It continues like this with a number in ascending order up to 15, and the following two lines after each number are a combination of four a's or b's separated by commas. I have tied this file to an array #diplo so I can specify specific lines of it.
I also have got a file that contains two columns of data with headers that I have converted into a hash of arrays (with each of the two columns being an array). The name of the hash is $lookup and the array names are the names of the headings. The actual arrays only start from the first value in each column that isn't a heading. This file looks like this:
haplo frequency
"|5,a,b,a,a|" 0.202493719
"|2,b,b,b,a|" 0.161139191
"|3,b,b,b,a|" 0.132602458
This file contains all of the possible combinations of a or b at the four positions combined with all numbers 0-14 and their associated frequencies. In other words, it includes all possible combinations from "|0,a,a,a,a|" followed be "|1,a,a,a,a|" through to "|13,b,b,b,b|" and "|14,b,b,b,b|".
I want my Perl code to go through each of the combinations of letters in #diplo starting with a,a,b,a and record the frequency associated with the row of the haplo array containing each number from 0-14, e.g. first recording the frequency associated with "|0,a,a,b,a|" then "|1,a,a,b,a|" etc.
The output would hopefully look like this:
0 #this is the number in the #diplo file and they increase in order from 0 up to 15
0.011 0.0023 0.003 0.0532 0.163 0.3421 0.128 0.0972 0.0869 0.05514 0.0219 0.0172 0.00824 0.00886 0.00196 #these are the frequencies associated with x,a,a,b,a where x is any number from 0 to 14.
My code:
And here is the Perl code I created to hopefully sort this out (there is more to create the arrays and such which I can post if required, but I didn't want to post a load of code if it isn't necessary):
my $irow = 1; #this is the row/element number in #diplo
my $lrow = 0; #this is the row/element in $lookup{'haplo'}
my $copynumber = 0;
#print "$copynumber, $diplo[2]";
while ($irow < $diplolines - 1) {
while ($copynumber < 15) {
while ($lrow < $uplines - 1) {
if ("|$copynumber,$diplo[$irow]|" = $lookup{'haplo'}[$lrow]) { ##this is the only line that causes errors
if ($copynumber == 0) {
print "$diplo[$irow-1]\n";
#print "$lookup{'frequency'}[$lrow]\t";
}
print "$lookup{'frequency'}[$lrow]\t";
}
$lrow = $lrow + 1;
}
$lrow = 0;
$copynumber = $copynumber + 1;
}
$lrow = 0;
$copynumber = 0;
$irow = $irow + 1;
}
However, the line if ("|$copynumber,$diplo[$irow]|" = $lookup{'haplo'}[$lrow]) is causing an error Can't modify string in scalar assignment near "]) ".
I have tried adding in speech marks, rounded brackets and apostrophes around various elements in this line but I still get some sort of variant on this error. I'm not sure how to get around this error.
Apologies for the long question, any help would be appreciated.
EDIT: Thanks for the suggestions regarding eq, it gets rid of the error and I now know a bit more about Perl than I did. However, even though I don't get an error now, if I put anything inside the if loop for this line, e.g. printing a number, it doesn't get executed. If I put the same command within the while loop but outside of the if, it does get executed. I have strict and warnings on. Any ideas?
= is assignment, == is numerical comparison, eq is string comparison.
You can't modify a string:
$ perl -e 'use strict; use warnings; my $foo="def";
if ("abc$foo" = "abcdef") { print "match\n"; } '
Found = in conditional, should be == at -e line 1.
Can't modify string in scalar assignment at -e line 1, near ""abcdef") "
Execution of -e aborted due to compilation errors.
Nonnumerical strings act like zeroes in a numerical comparison:
$ perl -e 'use strict; use warnings; my $foo="def";
if ("abc$foo" == 0) { print "match\n"; } '
Argument "abcdef" isn't numeric in numeric eq (==) at -e line 1.
match
A string comparison is probably what you want:
$ perl -e 'use strict; use warnings; my $foo="def";
if ("abc$foo" eq "abcdef") { print "match\n"; } '
match
This is the problematic expression:
"|$copynumber,$diplo[$irow]|" = $lookup{'haplo'}[$lrow]
The equals sign (=) is an assignment operator. It assigns the value on its right-hand side to the variable on its left-hand side. Therefore, the left-hand operand needs to be a variable, not a string as you have here.
I don't think you want to do an assignment here at all. I think you're trying to check for equality. So don't use an assignment operator, use a comparison operator.
Perl has two equality comparison operators. == does a numeric comparison to see if its operands are equal and eq does a string comparison. Why does Perl need two operators? Well Perl converts automatically between strings and numbers so it can't possibly know what kind of comparison you want to do. So you need to tell it.
What's the difference between the two types of comparison? Well, consider this code.
$x = '0';
$y = '0.0';
Are $x and $y equal? Well it depends on the kind of comparison you do. If you compare them as numbers then, yes, they are the same value (zero is the same thing whether it's an integer or a real number). But if you compare them as strings, they are different (they're not the same length for a start).
So we now know the following
$x == $y # this is true as it's a numeric comparison
$x eq $y # this is false as it's a string comparison
So let's go back to your code:
"|$copynumber,$diplo[$irow]|" = $lookup{'haplo'}[$lrow]
I guess you started with == here.
"|$copynumber,$diplo[$irow]|" == $lookup{'haplo'}[$lrow]
But that's not right as |$copynumber,$diplo[$irow]| is clearly as string, not a number. And Perl will give you a warning if you try to do a numeric comparison using a value that doesn't look like a number.
So you changed it to = and that doesn't work either as you've now changed it to an assignment.
What you really need is a string comparison:
"|$copynumber,$diplo[$irow]|" eq $lookup{'haplo'}[$lrow]
Here is the code:
my #s=<a b c d>;
for #s.kv {
for ($^k ... #s.elems) {
printf("%s ", $^v);
}
printf("\n");
}
Expected output is:
# a b c d
# b c d
# c d
# d
But it gives this error (possibly among others)
key 0, val 1 Too few positionals passed; expected 2 arguments but got 1
It looks like the positional variables of the main loop $^k and $^v can't be used inside the inner loop. How to fix it? Thanks.
Update: Typo inside inner loop fixed
So for what you want to do I'd approach it like this :
my #s = <a b c d>;
for ^#s.elems -> $start-index {
for #s[$start-index..*] -> $value {
printf("%s ", $value );
}
print("\n");
}
Though really I'd do this.
my #s = <a b c d>;
(^#s.elems).map( { #s[$_..*].join(" ").say } )
Get the range from 0 to the number of elements in the array. Then the slice from there to the end for each, join on spaces and say.
A note on variables like $^k these are scoped to the current block only (hence why your above code is not working). Generally you only really want to use them in map, grep or other such things. Where possible I'd always advise naming your variables, this makes them scoped inside inner blocks as well.
Scimon Proctor's answer is essentially correct, but I'll try to explain why your example does not work. For starters, kv returns "an interleaved sequence of indexes and values", so this:
my #s=<a b c d>;
.say for #s.kv;
prints
0
a
1
b
2
c
3
d
Essentially, you're doing one turn of the loop for every key and value. Grouping them in pairs using rotor might be closer to what you're looking for:
.say for #s.kv.rotor(2)
which will return:
(0 a)
(1 b)
(2 c)
(3 d)
Since with this we got the value couple with the index, we can do...
my #s=<a b c d>;
for #s.kv.rotor(2) -> ($k, $) {
"{#s[$_]} ".print for ($k..^#s.elems);
printf("\n");
}
Please note that there was also an error in the inner loop, whose range went beyond the actual indices in #s. But, again, Scimon's answer that uses maps is much shorter, idiomatic and straightforward. This one is just kind of dwimming your original program. As a matter of fact, we are throwing away the values, so this would actually be:
my #s=<a b c d>;
for #s.keys -> $k {
"{#s[$_]} ".print for ($k..^#s.elems);
printf("\n");
}
No need to use kv at all, and just make do with the keys.
I was playing around in Lua with a simple 'isPrime' function I made, and, ignoring the actual 'isPrime' function, which is irrelevant to this query, wrote the following code:
out = {}
for i = -10,20 do
out[i] = isPrime(i)
end
for k,v in ipairs(out) do
print(k,v)
end
My expectation was that the program would print every single key and its respective value, -10 through 20, but found instead that only 1 through 20 were printed. -10 through 0 were in the table, I found, after checking specifically for those key-value pairs, but oddly, they were never printed.
Can anyone explain why this happened? I feel I do not fully understand how Lua iterates and accesses its keys through the ipairs() function.
ipairs(t) will iterate over the key–value pairs (1,t[1]), (2,t[2]), ..., up to the first nil value. That's not what you want. Just use the style of your first loop
for i = -10,20 do
print(i, out[i])
end
I just got started with Clojure on Heroku, first reading this introduction.
Now in the phase of getting my hands dirty, I am facing this issue handling a database in a loop.
This is working:
(for
[s (db/query (env :database-url)
["select * from My_List"])]
; here one can do something with s, for example:
; print out (:field s)
)
But it is not enough to update variables inside the loop as I want.
Reading on the subject, I understand that Clojure having its own way of handling variables I need to use a loop pattern.
Here is what I tried:
(loop [a 0 b 1
s (db/query (env :database-url)
["select * from My_List"])]
; here I want to do something with s, for example
; print out (:field s)
; and do the following ... but it does not work!
(if (> (:otherField s) 5)
(:otherField s)
(recur (+ a (:otherField s)) b s))
)
Since I tried various ways of doing before writing this post, I know that the code above works except for the fact that I am doing something wrong concerning the database.
So here comes my question: What do I need to change to make it work?
I see, that it's hard to get to the functional thinking at first, when you're used to a different paradigm.
I don't think there is a correct explanation on “how to do this loop right”, because it's not right to do a loop here.
The two things that feel most incorrect to me:
Never do a SELECT * FROM table. This is not how relational databases are ment to be used. For example when you want the sum of all values greater than 5 you should do: SELECT SUM(field) FROM my_list WHERE field > 5
Don't think in loops (how to do it) but in what you want to do with the data:
I want to work on field :otherFIeld
I am only interested in values bigger than 5
I want the sum of all the remaining values
Then you come to something like this:
(reduce +
(filter #(> % 5)
(map :otherField
(db/query (env :database-url) ["select * from My_List"]))))
(No loop at all.)
How does Python know what "i" is when it is not defined, shouldn't there be an error? Probably a simple explanation, but I am new to learning Python.
def doubles (sum):
return sum * 2
myNum = 2
for i in range (0,3):
myNum = doubles(myNum)
print (myNum)
Haha :-) People are marking down your question, but I know that is one question must have came in every person's mind. Specially those who learned Python through Online courses and not through a teacher in person.
Well let me explain that in layman's term,
The method that you used is specially used for 1) lists and 2) lists within lists.
For eg,
example1= ['a','b','c'] # This is a simple list
example2 = [['a','b','c'],['a','b','c'],['a','b','c']] # This is list within lists.
Now, 'a','b' & 'c' are items in list.
So by saying,
for i in example1:
print i
we are actually saying,
for item in the list(example1):
print item
-------------------------
People use 'i', probably taken as abbreviation to item, or something else.
I don't know the history.
But, the fact is that, we can use anything instead or 'i' and Python will still consider it as an item in list.
Let me give you examples again.
example1= ['a','b','c'] # This is a simple list
example2 = [['a','b','c'],['a','b','c'],['a','b','c']] # This is list within lists.
for i in example1:
print i
[out]: a
b
c
now in example2, items are lists within lists. --- also, now i will use the word 'item' instead of 'i' --- the results regardless would be the same for both.
for item in example2:
print item
[out]: ['a','b','c']
['a','b','c']
['a','b','c']
people also use singulars and plurals to remember things,
so lets we have a list of alphabet.
letters=['a','b','c','d']
for letter in letters:
print letter
[out]: a
b
c
d
Hope that helps. There is much more to explain.
Keep researching and keep learning.
Regards,
Md. Mohsin
Using a variable as a loop control variable does assign to it each time through the loop.
As to "what it is"... Python is dynamically typed. The only thing it "is" is a name, just like any other variable.
i is assigned the value in the loop itself, it has no value (it is not defined) before the Python interpreter reaches the for line.
Its similar to how other for loops define variables. In C++ for example:
for(int i=0; i<5; i++){
cout << i << endl;
}
Here the variable i is only exists once the for loop is called.
i is assigned a value when the for loop runs, so the Python interpreter will not raise an error when the loop is run
long story short it creates a new variable without having to be defined and its value is whatever number your loop is on, for example if you had written:
num = 0
for i in range(3):
print(num)
num = num + 1
so for the first time this loop ran 'i' would equal 0 (because python lists/loops etc always start on 0 not 1), the second time it would equal 1, etc. and the 'num' you can ignore it's just an example of code you could have in a loop which would print out numbers in ascending order.
Levin