Develop Reference

Foobar - Test cases not passing - Disorderly escape

0/10 test cases are passing.
Here is the challenge description:
(to keep formatting nice - i put the description in a paste bin)
Link to challenge description: https://pastebin.com/UQM4Hip9
Here is my trial code - PYTHON - (0/10 test cases passed)
from math import factorial
from collections import Counter
from fractions import gcd
def cycle_count(c, n):
cc=factorial(n)
for a, b in Counter(c).items():
cc//=(a**b)*factorial(b)
return cc
def cycle_partitions(n, i=1):
yield [n]
for i in range(i, n//2 + 1):
for p in cycle_partitions(n-i, i):
yield [i] + p
def solution(w, h, s):
grid=0
for cpw in cycle_partitions(w):
for cph in cycle_partitions(h):
m=cycle_count(cpw, w)*cycle_count(cph, h)
grid+=m*(s**sum([sum([gcd(i, j) for i in cpw]) for j in cph]))
return grid//(factorial(w)*factorial(h))
print(solution(2, 2, 2)) #Outputs 7
This code works in my python compiler on my computer but not on the foobar challenge??
Am I losing accuracy or something?

NOT HELPFUL
Just a guess: wrong types of the return values of the functions? str vs. int?
Improved Answer
First a more detailed explanation for the 'type of values' thing:
In python, values are typed also if you do not have to declare a type. E.g.:
>>> i = 7
>>> s = '7'
>>> print(i, s, type(i), type(s), i == s)
7 7 <class 'int'> <class 'str'> False
I do not know the exact tests that are applied to the code, but typically those involve some equality test with ==. If the type of the expected value and those value returned does not match, the equality fails. Perhaps Elegant ways to support equivalence ("equality") in Python classes might help.
The instruction to the challenge (see pastebin) also explicitly mentions the expected return types: the returned values must be string (see line 41 / 47).
In addtion, the instruction also states that one should 'write a function answer(w, h, s)' (line 6). The posted solution implements a function solution.

I was also stuck on this problem.
The error is because of data type mismatch i.e your function returns an integer but the question specifically asked for a string.
Wrapping the return statement in str() will solve the problem.

Your return type is an integer. Convert it to a string, and it will work.
return str(grid//(factorial(w)*factorial(h)))

Python IP to integer conversion not working as expected

I am using the answer provided here to convert string IP to integer. But I am not getting expected output. Specifically, the method is
>>> ipstr = '1.2.3.4'
>>> parts = ipstr.split('.')
>>> (int(parts[0]) << 24) + (int(parts[1]) << 16) + \
(int(parts[2]) << 8) + int(parts[3])
but when I provide it the value 172.31.22.98 I get 2887718498 back. However, I expect to see value -1407248798 as provided by Google's Guava library https://google.github.io/guava/releases/20.0/api/docs/com/google/common/net/InetAddresses.html#fromInteger-int-
Also, I've verified that this service provides expected output but all of the answers provided by the aforementioned StackOverflow answer return 2887718498
Note that I cannot use any third party library. So I am pretty much limited to using a hand-written code (no imports)

A better way is to use the library method
>>> from socket import inet_aton
>>> int.from_bytes(inet_aton('172.31.22.98'), byteorder="big")
2887718498
This is still the same result you had above
Here is one way to view it as a signed int
>>> from ctypes import c_long
>>> c_long(2887718498)
c_long(-1407248798)
To do it without imports (but why? The above is all first party CPython)
>>> x = 2887718498
>>> if x >= (1<<31): x-= (1<<32)
>>> x
-1407248798

Found this post whilst trying to do the same as OP. Developed the below for my simple mind to understand and for someone to benefit from.
baseIP = '192.168.1.0'
baseIPFirstOctet = (int((baseIP).split('.')[0]))
baseIPSecondOctet = (int((baseIP).split('.')[1]))
baseIPThirdOctet = (int((baseIP).split('.')[2]))
baseIPFourthOctet = (int((baseIP).split('.')[3]))

TypeError: 'int' object is not callable on line 2

I am really new to Python so i was going through the interactive guide on runestone and entered this code:
t = int(input("number of yrs plsz"))
a = 10000(1+(0.08/12))**12t
print(a)
I am receiving this error
TypeError: 'int' object is not callable on line 2

10000() is syntax for calling a function (myfunction()). Instead, multiply with the * operator.
By using the parentheses, you are attempting to call the function 10000 which cannot be done because 10000 is an integer, not a function.

You can't multiply with parentheses, you need to explicitly type all the multiplication operators like so:
a = 10000 * (1 + (0.08 / 12)) ** (12 * t) # basically you can't call 10000() as a func

You cannot use the notation you're using on line 2.
When you write a = 1000(something), python thinks you're trying to call the method named "1000".
Rewrite as
a = 10000 * (1+(0.08/12))**12t)
Also I assume you're trying to do 12*t, you cannot just say 12t. Thats invalid syntax as well.
So really should be
a = 10000 * (1+(0.08/12))**12*t
I'm not sure on the math aspect of re-writing it this way, but at least it will run.

I think what you wanted to write is:
a = 10000 * (1 + (0.08 / 12)) ** (12 * t)
This will work like a charm, but the origin of the error might confuse you. The problem is that as soon as Python sees a parenthesis after any object, say
obj(p1)
it makes a call to the object's __call__ method like this
obj.__call__(p1)
If it is not defined you get the error you had. So what happened in your code is that Python reached the part a = 10000(..., it tried to do something like this
10000.__call__(1 + (0.08 / 12))
as 10000 is an instance of int class, it tried to find definition of __call__ method for that class. As you would guess, int class does not define __call__ method, thus the error.

Try this -
a = 10000*(1+(0.08/12))**12*t.
If you are multiplying t to 12, put bracket.

Strange Problem with RPy2

After installing RPy2 from
http://rpy.sourceforge.net/rpy2.html
I'm trying to use it in Python 2.6 IDLE but I'm getting this error:
>>> import rpy2.robjects as robjects
>>> robjects.r['pi']
<RVector - Python:0x0121D8F0 / R:0x022A1760>
What I'm doing wrong?

Have you tried looking at the vector that's returned?
>>> pi = robjects.r['pi']
>>> pi[0]
3.14159265358979

To expand on Shane's answer. rpy2 uses the following Python objects to represent the basic R types:
RVector: R scalars and vectors, R Lists are represented as RVectors with names, see below
RArray: an R matrix, essentially the RVector with a dimension
RDataFrame: an R data.frame
To coerce back to basic Python types look here.
As an example, I use this to convert an R List to a python dict:
rList = ro.r('''list(name1=1,name2=c(1,2,3))''')
pyDict = {}
for name,value in zip([i for i in rList.getnames()],[i for i in rList]):
if len(value) == 1: pyDict[name] = value[0]
else: pyDict[name] = [i for i in value]

In the Python interactive interpreter if an expression returns a value then that value is automatically printed. For example if you create a dictionary and extract a value from it the value is automatically printed, but if this was in an executing script this would not be the case. Look at the following simple example this is not an error but simply python printing the result of the expression:
>>> mymap = {"a":23}
>>> mymap["a"]
23
The same code in a python script would produce no output at all.
In your code you are accessing a map like structure with the code:
>>> robjects.r['pi']
This is returning some R2Py object for which the default string representation is: <RVector - Python:0x0121D8F0 / R:0x022A1760>
If you changed the code to something like:
pi = robjects.r['pi']
you would see no output but the result of the call (a vector) will be assigned to the variable pi and be available for you to use.
Looking at the R2Py documentation It seems many of the objects are by default printed as a type in <> brackets and some memory address information.

This is not an error, it's simply the 'repr' of the returned robject:
>>> r['pi']
<RVector - Python:0x2c14bd8 / R:0x3719538>
>>> repr(r['pi'])
'<RVector - Python:0x4b77908 / R:0x3719538>'
>>> str(r['pi'])
'[1] 3.141593'
>>> print r['pi']
[1] 3.141593
You can get the value of 'pi' accessing it by index
>>> r['pi'][0]
3.1415926535897931
To access element of named lists (the 'object$attribute' R syntax) I use
>>> l = r.list(a=r.c(1,2,3), b=r.c(4,5,6))
>>> print l
$a
[1] 1 2 3
$b
[1] 4 5 6
>>> print dict(zip(l.names, l))['a']
[1] 1 2 3
but I think there must be a better solution...

I found this as the only sensible, short discussion of how to go back and forth from R objects and python. naufraghi's solution prompted the following approach to converting a data.frame, which retains the nicer slicing capabilities of the dataframe:
In [69]: import numpy as np
In [70]: import rpy2.robjects as ro
In [71]: df = ro.r['data.frame'](a=r.c(1,2,3), b=r.c(4.0,5.0,6.3))
In [72]: df
Out[72]: <RDataFrame - Python:0x5492200 / R:0x4d00a28>
In [73]: print(df)
a b
1 1 4.0
2 2 5.0
3 3 6.3
In [74]: recdf = np.rec.fromarrays(df, names=tuple(df.names))
In [75]: recdf
Out[75]:
rec.array([(1, 4.0), (2, 5.0), (3, 6.2999999999999998)],
dtype=[('a', '<i4'), ('b', '<f8')])
Seems a bit off-topic at this point, but I'm not sure what the appropriate procedure would be to capture this question & answer of mine!

How to find number of bytes taken by python variable

Is there anyway i can know how much bytes taken by particular variable in python. E.g; lets say i have
int = 12
print (type(int))
it will print
<class 'int'>
But i wanted to know how many bytes it has taken on memory? is it possible?

You can find the functionality you are looking for here (in sys.getsizeof - Python 2.6 and up).
Also: don't shadow the int builtin!
import sys
myint = 12
print(sys.getsizeof(myint))

if you want to know size of int, you can use struct
>>> import struct
>>> struct.calcsize("i")
4
otherwise, as others already pointed out, use getsizeof (2.6). there is also a recipe you can try.

In Python >= 2.6 you can use sys.getsizeof.

Numpy offers infrastructure to control data size. Here are examples (py3):
import numpy as np
x = np.float32(0)
print(x.nbytes) # 4
a = np.zeros((15, 15), np.int64)
print(a.nbytes) # 15 * 15 * 8 = 1800
This is super helpful when trying to submit data to the graphics card with pyopengl, for example.

You could also take a look at Pympler, especially its asizeof module, which unlike sys.getsizeof works with Python >=2.2.

on python command prompt, you can use size of function
$ import python
$ import ctypes
$ ctypes.sizeof(ctypes.c_int)
and read more on it from https://docs.python.org/2/library/ctypes.html

In Python 3 you can use sys.getsizeof().
import sys
myint = 12
print(sys.getsizeof(myint))

The best library for that is guppy:
import guppy
import inspect
def get_object_size(obj):
h = guppy.hpy()
callers_local_vars = inspect.currentframe().f_back.f_locals.items()
vname = "Constant"
for var_name, var_val in callers_local_vars:
if var_val == obj:
vname = str(var_name)
size = str("{0:.2f} GB".format(float(h.iso(obj).domisize) / (1024 * 1024)))
return str("{}: {}".format(vname, size))

The accepted answer sys.getsizeof is correct.
But looking at your comment about the accepted answer you might want the number of bits a number is occupying in binary. You can use bit_length
(16).bit_length() # '10000' in binary
>> 5
(4).bit_length() # '100' in binary
>> 3