I'm trying to make a program that print every 100K-th odd prime number until 10M using Potion, my code:
last = 3
res = (last) # create array
loop:
last += 2
prime = true
len = res length -1
i = 0
while(i<len):
v = res at(i)
if(v*v > last): break.
if(last%v == 0): prime = false, break.
i += 1
.
if(prime):
res append(last)
if(res size % 100000 == 0): last print.
if(last>9999999): break.
.
.
But this gives Segmentation fault (core dumped), I wonder what's wrong?
For reference, the working Ruby version:
res = [last=3]
loop do
last += 2
prime = true
res.each do |v|
break if v*v > last
if last % v == 0
prime = false
break
end
end
if prime
res << last
puts last if res.length % 100000 == 0
break if last > 9999999
end
end
The output should be:
1299721
2750161
4256249
5800139
7368791
8960467
and no, this is not a homework, just out of curiosity.
you found it out by yourself, great!
println is called say in potion.
And it crashed in res size.
E.g. use this for debbugging:
rm config.inc
make DEBUG=1
bin/potion -d -Dvt example/100thoddprime.pn
and then press enter until you get to the crash.
(example/100thoddprime.pn:18): res append(last)
>
; (3, 5)
[95] getlocal 1 1 ; (3, 5)
[96] move 2 1 ; (3, 5)
[97] loadk 1 5 ; size
[98] bind 1 2 ; function size()
[99] loadpn 3 0 ; nil
[100] call 1 3Segmentation fault
so size on res returned nil, and this caused the crash.
And instead of last print, "\n" print.
Just do last say.
This came from perl6 syntax, sorry :)
My bad, I forgot to change from res length -1 to res length when changing from 0 to len (i), because this syntax not recognized as a loop (failed to receive break).
last = 3
res = (last)
loop:
last println
last += 2
prime = true
len = res length
i = 0
while(i<len):
v = res at(i)
if(v*v > last): break.
if(last%v == 0): prime = false, break.
i += 1
.
if(prime):
res append(last)
if(res length % 100000 == 0): last print, "\n" print.
if(last>9999999): break.
.
.
Related
So i have for example t=[1.0, 1.0, 1.6, 1.125, 1.5]
I want to print the indices for the elements that have the minimum value min(t) but i want them to start from 1
So for this example i want to print 1 2
It's working when i do this:
for j in range(len(t)):
if t[j]==min(t):
print j+1,`
Output :
1 2
But it's not working with this:
for j in t:
if j==min(t):
a=t.index(j)
print a+1,`
Output :
1 1
Why is that?
Your error is here
a = t.index(j)
t.index(j) is going to give you the first index of that value.
You can make this:
m = min(t)
cont = 0
for j in t:
if j == m:
print cont+1,
cont += 1
Or
for idx, val in enumerate(t):
if val == m:
print idx + 1,
I have this list: l = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20] and I would like to iterate on it, printing something like
0: [1,2,3,4]
1: [2,3,4,5]
2: [3,4,5,6]
...
n: [17,18,19,20]
So far I made this code to print 5 elements at a time, but the last iteration prints 3:
for index, item in enumerate(l):
if index == 0 or index == 1 or index == 2:
continue
print index, l[index - 3:index + 2]
How can I solve this?
You're on the right track with your list slices. Here's a tweak to make it easier:
sub_len = 4
for i in range(len(mylist) - sub_len + 1):
print l[i:i + sub_len]
Where sub_len is the desired length of the slices you print.
demo
Suppose I have the following local macro:
loc a = 12.000923
I would like to get the decimal position of the first non-zero decimal (4 in this example).
There are many ways to achieve this. One is to treat a as a string and to find the position of .:
loc a = 12.000923
loc b = strpos(string(`a'), ".")
di "`b'"
From here one could further loop through the decimals and count since I get the first non-zero element. Of course this doesn't seem to be a very elegant approach.
Can you suggest a better way to deal with this? Regular expressions perhaps?
Well, I don't know Stata, but according to the documentation, \.(0+)? is suported and it shouldn't be hard to convert this 2 lines JavaScript function in Stata.
It returns the position of the first nonzero decimal or -1 if there is no decimal.
function getNonZeroDecimalPosition(v) {
var v2 = v.replace(/\.(0+)?/, "")
return v2.length !== v.length ? v.length - v2.length : -1
}
Explanation
We remove from input string a dot followed by optional consecutive zeros.
The difference between the lengths of original input string and this new string gives the position of the first nonzero decimal
Demo
Sample Snippet
function getNonZeroDecimalPosition(v) {
var v2 = v.replace(/\.(0+)?/, "")
return v2.length !== v.length ? v.length - v2.length : -1
}
var samples = [
"loc a = 12.00012",
"loc b = 12",
"loc c = 12.012",
"loc d = 1.000012",
"loc e = -10.00012",
"loc f = -10.05012",
"loc g = 0.0012"
]
samples.forEach(function(sample) {
console.log(getNonZeroDecimalPosition(sample))
})
You can do this in mata in one line and without using regular expressions:
foreach x in 124.000923 65.020923 1.000022030 0.0090843 .00000425 {
mata: selectindex(tokens(tokens(st_local("x"), ".")[selectindex(tokens(st_local("x"), ".") :== ".") + 1], "0") :!= "0")[1]
}
4
2
5
3
6
Below, you can see the steps in detail:
. local x = 124.000823
. mata:
: /* Step 1: break Stata's local macro x in tokens using . as a parsing char */
: a = tokens(st_local("x"), ".")
: a
1 2 3
+----------------------------+
1 | 124 . 000823 |
+----------------------------+
: /* Step 2: tokenize the string in a[1,3] using 0 as a parsing char */
: b = tokens(a[3], "0")
: b
1 2 3 4
+-------------------------+
1 | 0 0 0 823 |
+-------------------------+
: /* Step 3: find which values are different from zero */
: c = b :!= "0"
: c
1 2 3 4
+-----------------+
1 | 0 0 0 1 |
+-----------------+
: /* Step 4: find the first index position where this is true */
: d = selectindex(c :!= 0)[1]
: d
4
: end
You can also find the position of the string of interest in Step 2 using the
same logic.
This is the index value after the one for .:
. mata:
: k = selectindex(a :== ".") + 1
: k
3
: end
In which case, Step 2 becomes:
. mata:
:
: b = tokens(a[k], "0")
: b
1 2 3 4
+-------------------------+
1 | 0 0 0 823 |
+-------------------------+
: end
For unexpected cases without decimal:
foreach x in 124.000923 65.020923 1.000022030 12 0.0090843 .00000425 {
if strmatch("`x'", "*.*") mata: selectindex(tokens(tokens(st_local("x"), ".")[selectindex(tokens(st_local("x"), ".") :== ".") + 1], "0") :!= "0")[1]
else display " 0"
}
4
2
5
0
3
6
A straighforward answer uses regular expressions and commands to work with strings.
One can select all decimals, find the first non 0 decimal, and finally find its position:
loc v = "123.000923"
loc v2 = regexr("`v'", "^[0-9]*[/.]", "") // 000923
loc v3 = regexr("`v'", "^[0-9]*[/.][0]*", "") // 923
loc first = substr("`v3'", 1, 1) // 9
loc first_pos = strpos("`v2'", "`first'") // 4: position of 9 in 000923
di "`v2'"
di "`v3'"
di "`first'"
di "`first_pos'"
Which in one step is equivalent to:
loc first_pos2 = strpos(regexr("`v'", "^[0-9]*[/.]", ""), substr(regexr("`v'", "^[0-9]*[/.][0]*", ""), 1, 1))
di "`first_pos2'"
An alternative suggested in another answer is to compare the lenght of the decimals block cleaned from the 0s with that not cleaned.
In one step this is:
loc first_pos3 = strlen(regexr("`v'", "^[0-9]*[/.]", "")) - strlen(regexr("`v'", "^[0-9]*[/.][0]*", "")) + 1
di "`first_pos3'"
Not using regex but log10 instead (which treats a number like a number), this function will:
For numbers >= 1 or numbers <= -1, return with a positive number the number of digits to the left of the decimal.
Or (and more specifically to what you were asking), for numbers between 1 and -1, return with a negative number the number of digits to the right of the decimal where the first non-zero number occurs.
digitsFromDecimal = (n) => {
dFD = Math.log10(Math.abs(n)) | 0;
if (n >= 1 || n <= -1) { dFD++; }
return dFD;
}
var x = [118.8161330, 11.10501660, 9.254180571, -1.245501523, 1, 0, 0.864931613, 0.097007836, -0.010880074, 0.009066729];
x.forEach(element => {
console.log(`${element}, Digits from Decimal: ${digitsFromDecimal(element)}`);
});
// Output
// 118.816133, Digits from Decimal: 3
// 11.1050166, Digits from Decimal: 2
// 9.254180571, Digits from Decimal: 1
// -1.245501523, Digits from Decimal: 1
// 1, Digits from Decimal: 1
// 0, Digits from Decimal: 0
// 0.864931613, Digits from Decimal: 0
// 0.097007836, Digits from Decimal: -1
// -0.010880074, Digits from Decimal: -1
// 0.009066729, Digits from Decimal: -2
Mata solution of Pearly is very likable, but notice should be paid for "unexpected" cases of "no decimal at all".
Besides, the regular expression is not a too bad choice when it could be made in a memorable 1-line.
loc v = "123.000923"
capture local x = regexm("`v'","(\.0*)")*length(regexs(0))
Below code tests with more values of v.
foreach v in 124.000923 605.20923 1.10022030 0.0090843 .00000425 12 .000125 {
capture local x = regexm("`v'","(\.0*)")*length(regexs(0))
di "`v': The wanted number = `x'"
}
I have a beginner question. Loops are extremely hard for me to understand, so it's come to me asking for help.
I am trying to create a function to count the amount of even numbers in a user input list, with a negative at the end to show the end of the list. I know I need to use a while loop, but I am having trouble figuring out how to walk through the indexes of the input list. This is what I have so far, can anyone give me a hand?
def find_even_count(numlist):
count = 0
numlist.split()
while numlist > 0:
if numlist % 2 == 0:
count += 1
return count
numlist = raw_input("Please enter a list of numbers, with a negative at the end: ")
print find_even_count(numlist)
I used the split to separate out the indexes of the list, but I know I am doing something wrong. Can anyone point out what I am doing wrong, or point me to a good step by step explanation of what to do here?
Thank you guys so much, I know you probably have something more on your skill level to do, but appreciate the help!
You were pretty close, just a couple of corrections:
def find_even_count(numlist):
count = 0
lst = numlist.split()
for num in lst:
if int(num) % 2 == 0:
count += 1
return count
numlist = raw_input("Please enter a list of numbers, with a negative at the end: ")
print find_even_count(numlist)
I have used a for loop rather than a while loop, stored the outcome of numlist.split() to a variable (lst) and then just iterated over this.
You have a couple of problems:
You split numlist, but don't assign the resulting list to anything.
You then try to operate on numlist, which is still the string of all numbers.
You never try to convert anything to a number.
Instead, try:
def find_even_count(numlist):
count = 0
for numstr in numlist.split(): # iterate over the list
num = int(numstr) # convert each item to an integer
if num < 0:
break # stop when we hit a negative
elif num % 2 == 0:
count += 1 # increment count for even numbers
return count # return the total
Or, doing the whole thing in one line:
def find_even_count(numlist):
return sum(num % 2 for num in map(int, numlist.split()) if num > 0)
(Note: the one-liner will fail in cases where the user tries to trick you by putting more numbers after the "final" negative number, e.g. with numlist = "1 2 -1 3 4")
If you must use a while loop (which isn't really the best tool for the job), it would look like:
def find_even_count(numlist):
index = count = 0
numlist = list(map(int, numlist.split()))
while numlist[index] > 0:
if numlist[index] % 2 == 0:
count += 1
index += 1
return count
I'm working through the book NLP with Python, and I came across this example from an 'advanced' section. I'd appreciate help understanding how it works. The function computes all possibilities of a number of syllables to reach a 'meter' length n. Short syllables "S" take up one unit of length, while long syllables "L" take up two units of length. So, for a meter length of 4, the return statement looks like this:
['SSSS', 'SSL', 'SLS', 'LSS', 'LL']
The function:
def virahanka1(n):
if n == 0:
return [""]
elif n == 1:
return ["S"]
else:
s = ["S" + prosody for prosody in virahanka1(n-1)]
l = ["L" + prosody for prosody in virahanka1(n-2)]
return s + l
The part I don't understand is how the 'SSL', 'SLS', and 'LSS' matches are made, if s and l are separate lists. Also in the line "for prosody in virahanka1(n-1)," what is prosody? Is it what the function is returning each time? I'm trying to think through it step by step but I'm not getting anywhere. Thanks in advance for your help!
Adrian
Let's just build the function from scratch. That's a good way to understand it thoroughly.
Suppose then that we want a recursive function to enumerate every combination of Ls and Ss to make a given meter length n. Let's just consider some simple cases:
n = 0: Only way to do this is with an empty string.
n = 1: Only way to do this is with a single S.
n = 2: You can do it with a single L, or two Ss.
n = 3: LS, SL, SSS.
Now, think about how you might build the answer for n = 4 given the above data. Well, the answer would either involve adding an S to a meter length of 3, or adding an L to a meter length of 2. So, the answer in this case would be LL, LSS from n = 2 and SLS, SSL, SSSS from n = 3. You can check that this is all possible combinations. We can also see that n = 2 and n = 3 can be obtained from n = 0,1 and n=1,2 similarly, so we don't need to special-case them.
Generally, then, for n ≥ 2, you can derive the strings for length n by looking at strings of length n-1 and length n-2.
Then, the answer is obvious:
if n = 0, return just an empty string
if n = 1, return a single S
otherwise, return the result of adding an S to all strings of meter length n-1, combined with the result of adding an L to all strings of meter length n-2.
By the way, the function as written is a bit inefficient because it recalculates a lot of values. That would make it very slow if you asked for e.g. n = 30. You can make it faster very easily by using the new lru_cache from Python 3.3:
#lru_cache(maxsize=None)
def virahanka1(n):
...
This caches results for each n, making it much faster.
I tried to melt my brain. I added print statements to explain to me what was happening. I think the most confusing part about recursive calls is that it seems to go into the call forward but come out backwards, as you may see with the prints when you run the following code;
def virahanka1(n):
if n == 4:
print 'Lets Begin for ', n
else:
print 'recursive call for ', n, '\n'
if n == 0:
print 'n = 0 so adding "" to below'
return [""]
elif n == 1:
print 'n = 1 so returning S for below'
return ["S"]
else:
print 'next recursivly call ' + str(n) + '-1 for S'
s = ["S" + prosody for prosody in virahanka1(n-1)]
print '"S" + each string in s equals', s
if n == 4:
print '**Above is the result for s**'
print 'n =',n,'\n', 'next recursivly call ' + str(n) + '-2 for L'
l = ["L" + prosody for prosody in virahanka1(n-2)]
print '\t','what was returned + each string in l now equals', l
if n == 4:
print '**Above is the result for l**','\n','**Below is the end result of s + l**'
print 'returning s + l',s+l,'for below', '\n','='*70
return s + l
virahanka1(4)
Still confusing for me, but with this and Jocke's elegant explanation, I think I can understand what is going on.
How about you?
Below is what the code above produces;
Lets Begin for 4
next recursivly call 4-1 for S
recursive call for 3
next recursivly call 3-1 for S
recursive call for 2
next recursivly call 2-1 for S
recursive call for 1
n = 1 so returning S for below
"S" + each string in s equals ['SS']
n = 2
next recursivly call 2-2 for L
recursive call for 0
n = 0 so adding "" to below
what was returned + each string in l now equals ['L']
returning s + l ['SS', 'L'] for below
======================================================================
"S" + each string in s equals ['SSS', 'SL']
n = 3
next recursivly call 3-2 for L
recursive call for 1
n = 1 so returning S for below
what was returned + each string in l now equals ['LS']
returning s + l ['SSS', 'SL', 'LS'] for below
======================================================================
"S" + each string in s equals ['SSSS', 'SSL', 'SLS']
**Above is the result for s**
n = 4
next recursivly call 4-2 for L
recursive call for 2
next recursivly call 2-1 for S
recursive call for 1
n = 1 so returning S for below
"S" + each string in s equals ['SS']
n = 2
next recursivly call 2-2 for L
recursive call for 0
n = 0 so adding "" to below
what was returned + each string in l now equals ['L']
returning s + l ['SS', 'L'] for below
======================================================================
what was returned + each string in l now equals ['LSS', 'LL']
**Above is the result for l**
**Below is the end result of s + l**
returning s + l ['SSSS', 'SSL', 'SLS', 'LSS', 'LL'] for below
======================================================================
This function says that:
virakhanka1(n) is the same as [""] when n is zero, ["S"] when n is 1, and s + l otherwise.
Where s is the same as the result of "S" prepended to each elements in the resulting list of virahanka1(n - 1), and l the same as "L" prepended to the elements of virahanka1(n - 2).
So the computation would be:
When n is 0:
[""]
When n is 1:
["S"]
When n is 2:
s = ["S" + "S"]
l = ["L" + ""]
s + l = ["SS", "L"]
When n is 3:
s = ["S" + "SS", "S" + "L"]
l = ["L" + "S"]
s + l = ["SSS", "SL", "LS"]
When n is 4:
s = ["S" + "SSS", "S" + "SL", "S" + "LS"]
l = ["L" + "SS", "L" + "L"]
s + l = ['SSSS", "SSL", "SLS", "LSS", "LL"]
And there you have it, step by step.
You need to know the results of the other function calls in order to calculate the final value, which can be pretty messy to do manually as you can see. It is important though that you do not try to think recursively in your head. This would cause your mind to melt. I described the function in words, so that you can see that these kind of functions is are descriptions, and not a sequence of commands.
The prosody you see, that is a part of s and l definitions, are variables. They are used in a list-comprehension, which is a way of building lists. I've described earlier how this list is built.