istr-python 0.1.1

istr - strings you can count on

Introduction

The istr module has exactly one class: istr.

With this it is possible to interpret strings as if they were integers.

This can be very handy for solving puzzles, but also for other purposes. For instance the famous send more money puzzle

  S E N D
  M O R E
--------- +
M O N E Y

can be nicely, albeit not very efficient, coded as:

import itertools
from istr import istr

for s, e, n, d, m, o, r, y in istr(itertools.permutations(range(10), 8)):
    if m and ((s|e|n|d) + (m|o|r|e) == (m|o|n|e|y)):
        print(f" {s|e|n|d}")
        print(f" {m|o|r|e}")
        print("-----")
        print(f"{m|o|n|e|y}")

And it is a nice demonstration of extending a class (str) with extra and changed functionality.

Installation

Installing istr with pip is easy.

$ pip install istr-python

or when you want to upgrade,

$ pip install istr-python --upgrade

Alternatively, istr.py can be just copied into you current work directory from GitHub (https://github.com/salabim/istr).

No dependencies!

Usage

Just start with

from istr import istr

Now we can define some istrs:

four = istr("4")
five = istr("5")

Then we can do

x= four * five

, after which x is istr("20")

And now we can do

print(x == 20)
print(x == "20")

resulting in two times True. That's because istrs instances are treated as int, although they are strings.

That means that we can also say

print(x < 30)
print(x >= "10")

again resulting in two times True.

In contrast to an ordinary string

print(four + five)

prints 9, as istr are treated as ints.

Please note that four and five could have also be initialized with

four = istr(4)
five = istr(5)

or even

four, five = istr(4, 5)

But how can we concatenate istrs? Just use the or operator (|):

print(four | five)

will output 45.

And the result is again an istr.

That means that

(four | five) / 3

is istr("9").

In order to repeat a string in the usual sense, you cannot use 3 * four, as that woud be 12.

We use the matrix multiplication operator (@) for this. So 3 @ four is 444. As is four @ 3.

Also allowed are

abs(four)
-four 

The bool operator works on the integer value of an istr. So bool("0") ==> False bool("1") ==> True The code

if istr("0"):
    print("True")
else:
    print("False")

will print False

For the in operator, an istr is treated as an ordinary string, although it is possible to use ints as well:

"34" in istr(1234)
34 in istr(1234)

On the left hand side an istr is always treated as a string:

istr(1234) in "01234566890ABCDEF"

Note that all calculations are strictly integer calculations. That means that if a float variale is ever produced it will be converted to an int. Also divisions are always floor divisions!

There's a special case for istr(""). This is a proper empty string, but also represents the value of 0. That is to allow for istr("").join(i for i in "01234)", resulting in istr("01234").

Sorting a list of istrs is based on the integer value, not the string. So

" ".join(sorted("1 3 2 4 5 6 11 7 9 8 10 12 0".split()))

is

"0 1 10 11 2 3 4 5 6 7 8 9"

,whereas

" ".join(sorted(istr("1 3 2 4 5 6 11 7 9 8 10 12 0".split())))

is

"0 1 2 3 4 5 6 7 8 9 10 11"

Using other values for istr than numeric value or str

Apart from with simple numeric (to be interpreted as an int) or str, istr can be initialized with several other types:

• if a dict (or subtype of dict), the same type dict will be returned with all values istr'ed

  istr({0: 0, 1: 1, 2: 4}) ==> {0: istr("0"), 1: istr("1"), 2: istr("4")}

• if an iterator, the iterator will be mapped with istr

  istr(i * i for i in range(3)) ==> <map object>

  list(istr(i * i for i in range(3))) ==> [istr("0"), istr("1"), istr("4")]

• if an iterable, the same type will be returned with all elements istr'ed

  istr([0, 1, 4]) ==> [istr("0"), istr("1"), istr("4")]

  istr((0, 1, 4)) ==> (istr("0"), istr("1"), istr("4"))

  istr({0, 1, 4}) ==> {istr("4"), istr("0"), istr("1")} # or similar

• if a range, an istr.range instance will be returned

  istr(range(3)) ==> istr.range(3)

  list(istr(range(3))) ==> [istr("0"), istr("1"), istr("2")]

  len(istr(range(3))) ==> 3

• if an istr.range instance, the same istr.range will be returned

More than one parameter for istr

It is possible to give more than one parameter, in which case a tuple of the istrs of the parameters will be returned, which can be handy to unpack multiple values, e.g.

a, b, c = istr(5, 6, 7) ==> a=istr("5") , b=istr("6"), c=istr("7")

test for even/odd

It is possible to test for even/odd with the

is_even and is_odd method, e.g.

print(istr(4).is_even())
print(istr(5).is_odd())

This will print True twice.

reverse an istr

The method istr.reversed() will return the an istr with the reversed content:

print(repr(istr(456).reversed()))
print(repr(istr("0456").reversed()))

result:

istr("654")
istr("6540")

The same can -of course- be achieved with

print(repr(istr(456)[::-1]))
print(repr(istr("0456")[::-1]))

Note that is impossible to reverse a negative istr.

enumerate with istrs

The istr.enumerate method can be used just as the builtin enumerate function. The iteration counter however is an istr rather than an int. E.g.

    for i,c in istr.enumerate("abc"):
        print(f"{repr(i)} {c}")

prints

istr('0') a
istr('1') b
istr('2') c

concatenate an iterable

The istr.concat1 method can be useful to map all items of an iterable to istr` and then concatenate these.

list(istr.concat(((1,2),(3,4))) ==> istr([12,34])

list(istr.concat(itertools.permutations(range(3),2))) ==> [istr('01'), istr('02'), istr('10'), istr('12'), istr('20'), istr('21')]

Subclassing istr

When a class is derived from istr, all methods will return that newly derived class.

E.g.

class jstr(istr):
    ...
    
print(repr(jstr(4) * jstr(5)))

will print jstr('20')

Changing the way repr works

It is possible to control the way an istr instance will be repr'ed.

By default, the istr('5') is represented as istr('5').

With the istr.repr_mode() context manager, that can be changed:

with istr.repr_mode("str"):
    five = istr('5')
    print(repr(five))
with istr.repr_mode("int"):
    five = istr('5')
    print(repr(five))
with istr.repr_mode("istr"):
    five = istr('5')
    print(repr(five))

This will print

'5'
5
istr('5')

Note that the way an istr is represented is determined at initialization.

It is also possible to set the repr mode without a context manager:

istr.repr_mode("str")
five = istr('5')
print(repr(five))

This will print

'5'

Finally, the current repr mode can be queried with istr.repr_mode(). So upon start:

print(repr(istr.repr_mode()))

will output istr.

Changing the base system

By default, istr works in base 10. However it is possible to change the base system with the istr.base() context manager / method.

Any base between 2 and 36 may be used.

Note that the integer is always stored in base 10 mode, but the string representation will reflect the chosen base (at time of initialization).

Some examples:

with istr.base(16):
    a = istr("7fff")
    print(int(a))

    b = istr(15)
    print(repr(b))

This will result in

32767
istr('F')

All calculations are done in the decimal 10 system.

Note that the way an istr is interpreted is determined at initialization.

It is also possible to set the repr mode without a context manager:

istr.base(16)
print(int(istr("7fff")))

This will print

32767

Finally, the current base can be queried with istr.base(), so upon start:

print(istr.base())

will result in 10.

Changing the format of the string

By default, istr does not change the way an istr is stored when a str is to initialize:

repr('4')) ==> istr('4')

repr(' 4')) ==> istr(' 4')

repr('4 ')) ==> istr('4 ')

For initializing with an int (or other numeric) value, the string is simply the str representation

repr(4)) ==> istr('4')

With the istr.format() context manager this behavior can be changed. If the format specifier is a number, most likely a single digit, that will be the minimum number of characters in the string:

with istr.format("3"):
    print(repr(istr(1)))
    print(repr(istr(12)))
    print(repr(istr(123)))
    print(repr(istr(1234)))

will print

istr('  1')
istr(' 12')
istr('123')
istr('1234')

If the string starts with a 0, the string will be zero filled:

with istr.format("03"):
    print(repr(istr(1)))
    print(repr(istr(12)))
    print(repr(istr(123)))
    print(repr(istr(1234)))

will print

istr('001')
istr('012')
istr('123')
istr('1234')

Note that if a format other than the default '' is used, the string will reformatted even if the istr is specified with a string:

with istr.format("03"):
    print(repr(istr("  12 ")))

will result in istr('0012')

Remark: For bases other than 10, the string will never be reformatted!

Test script

There's an extensive pytest script in the \tests directory.

This script also shows clearly the ways istr can be used.