What's wrong whith this code:
def Prime(a,b):
for n in range(a,b):
for x in range(a,n):
if n % x == 0:
print(n,"equales to",x,"*",n//x)
else:
break
else:
print(n,"is a prime number")
Prime(2,16)
I want to print all prime numbers in an interval, if not, print all possible divisions of that non-prime number
the output:
2 is a prime number
4 equales to 2 * 2
6 equales to 2 * 3
6 equales to 3 * 2
8 equales to 2 * 4
10 equales to 2 * 5
12 equales to 2 * 6
12 equales to 3 * 4
12 equales to 4 * 3
14 equales to 2 * 7
The for/else statement works in such a way that only if the loop is exhausted without reaching a break statement, the else will be executed.
This is the opposite of what you meant I am guessing.
What I would do, is simply create a "flag" variable to check at the end of the loop. Something like:
def Prime(a,b):
for n in range(a,b):
is_prime = True
for x in range(a,n):
if n % x == 0:
is_prime = False
print(n,"equales to",x,"*",n//x)
if is_prime:
print(n,"is a prime number")
This gives on Prime(2,16):
2 is a prime number
3 is a prime number
4 equales to 2 * 2
5 is a prime number
6 equales to 2 * 3
6 equales to 3 * 2
7 is a prime number
8 equales to 2 * 4
8 equales to 4 * 2
9 equales to 3 * 3
10 equales to 2 * 5
10 equales to 5 * 2
11 is a prime number
12 equales to 2 * 6
12 equales to 3 * 4
12 equales to 4 * 3
12 equales to 6 * 2
13 is a prime number
14 equales to 2 * 7
14 equales to 7 * 2
15 equales to 3 * 5
15 equales to 5 * 3
Related
Could someone explain the code? I just can not understand why this code gives output like this:
1
3
6
10
15
21
I expected the code to give something like this:
1
3
5
7
9
11
What am I missing here?
def tri_recursion(k):
if(k > 0):
result = k + tri_recursion(k-1)
print(result)
else:
result = 0
return result
tri_recursion(6)
For your recursive function, the termination condition is k=0.
It's clear that if k=0, tri_recursion(0) = 0.
If k=1, tri_recursion(1) = 1 + tri_recursion(0), which from above, is 1 + 0 or 1.
If k=2, tri_recursion(2) = 2 + tri_recursion(1), which from above, is 2 + 1 or 3.
If k=3, tri_recursion(3) = 3 + tri_recursion(2), which from above, is 3 + 3 or 6.
If k=4, tri_recursion(4) = 5 + tri_recursion(3), which from above, is 4 + 6 or 10.
If k=5, tri_recursion(5) = 4 + tri_recursion(4), which from above, is 5 + 10 or 15.
If k=6, tri_recursion(6) = 6 + tri_recursion(5), which from above, is 6 + 15 or 21.
See the pattern?
Your code is calculating the sum of numbers up to n where n is 6 in the above case. The print statement prints the intermediate results. Hence the output 1 3 6 10 15 21.
1 - The sum of numbers from 0 to 1
3 - The sum of numbers from 0 to 2
6 - The sum of numbers from 0 to 3
10 - The sum of numbers from 0 to 4
15 - The sum of numbers from 0 to 5
21 - The sum of numbers from 0 to 6
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
I need to create a incremental series for a given value of dataframe in python.
Any help much appreciated
Suppose I have dataframe column
df['quadrant']
Out[6]:
0 4
1 4
2 4
3 3
4 3
5 3
6 2
7 2
8 2
9 1
10 1
11 1
I want to create a new column such that
index quadrant new value
0 4 1
1 4 5
2 4 9
3 3 2
4 3 6
5 3 10
6 2 3
7 2 7
8 2 11
9 1 4
10 1 8
11 1 12
Using Numpy, you can create the array as:
import numpy as np
def value(q, k=1):
diff_quadrant = np.diff(q)
j = 0
ramp = []
for i in np.where(diff_quadrant != 0)[0]:
ramp.extend(list(range(i-j+1)))
j = i+1
ramp.extend(list(range(len(quadrant)-j)))
ramp = np.array(ramp) * k # sawtooth-shaped array
a = np.ones([len(quadrant)], dtype = np.int)*5
return a - q + ramp
quadrant = np.array([3, 3, 3, 3, 4, 4, 4, 2, 2, 1, 1, 1])
b = value(quadrant, 4)
# [ 2 6 10 14 1 5 9 3 7 4 8 12]
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}")
In J:
a =: 2 3 $ 1 2 3 4 5 6
Gives:
1 2 3
4 5 6
Which is a 2 3 shaped array.
If I do:
0 1 { a
I (noting that 0 1 is a 2 shaped list) expected to have back:
1 2 3 4 5 6
But got the following instead:
1 2 3
4 5 6
Reading the documentation I was expecting the shape of the index to kinda govern the shape of the answer.
Can someone clarify what I am missing here?
Higher-dimensional arrays may help make this clear. An array with n dimensions has items with n-1 dimensions. When you select an item from ({) a three-dimensional array, your result is a two-dimensional array:
1 { i. 5 3 4
12 13 14 15
16 17 18 19
20 21 22 23
When you select multiple items from an array, the items are assembled into a new array, using each atom of x to select a item of y. This might be where you picked up the idea that the shape of x affects the shape of the result.
2 1 0 2 { 'set'
test
$ 2 1 0 2
4
$ 'test'
4
The dimensions of the result is equal to the dimensions of x plus the dimensions of the items of y. So, if you have a two-dimensional x taking two-dimensional items from a three-dimensional y, you will have a four-dimensional result:
(2 2 $ 1 1 0 1) { i. 5 3 4
12 13 14 15
16 17 18 19
20 21 22 23
12 13 14 15
16 17 18 19
20 21 22 23
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
16 17 18 19
20 21 22 23
$ (2 2 $ 1 1 0 1) { i. 5 3 4
2 2 3 4
One final note: the monadic Ravel (,) will reduce the result to a list (one-dimensional array).
, 0 1 { 2 3 $ 1 2 3 4 5 6
1 2 3 4 5 6
, i. 2 2 2 2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
From ({) selects the items of a noun. For 2 3 $ 1 2 3 4 5 6 the items are the two rows because items are the components that make up the noun.
[ a=. 2 3 $ 1 2 3 4 5
1 2 3
4 5 1
0 { a
1 2 3
If you just had 1 2 3 then the items would be the individual atoms.
[ b=. 1 2 3
1 2 3
0 { b
1
If you used 1 3 $ 1 2 3 then there is only one item and the result would be
[ c=. 1 3 $ 1 2 3
1 2 3
0 { c
1 2 3
The number of items can be found with Tally (#), and is the lead dimension of the Shape ($) of the noun.
$ a
2 3
$ b
3
$ c
1 3
# a
2
# b
3
# c
1