istr-python 1.0.5

istr - strings you can count on

Introduction

The istr module makes it possible to use 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, where each letter has to be replaced by a unique digit (0-9)

  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}')

Of, if we want to add all the digits in a string:

sum_digits = sum(istr('9282334'))  # answer 31

And the module is a 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

Start

Just start with

from istr import istr

Use istr as int

We can define an istr:

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

The variables four and five can now be used as if they were int:

twenty = four * five

, after which x is istr('20')

The same can be done with

twenty = 4 * five

or

twenty = four * 5

And now twenty can be used as if it was an int as well. So

twenty - four

is istr('16')

We can do all the usual arithmetic operations on istrs, e.g.

-four + (twenty / 2)

is istr('6')

And we can test for equality. So:

twenty == 20

is True.

But istrs are also strings. So

twenty == '20'

is also True!

For the order comparisons (<=, <, >, >=), the istr is always interpreted as an int.

That means that

twenty < 30
twenty >= '10' # here '10' is converted to the integer 10 for the comparison

are bothTrue.

In contrast to an ordinary string

print(four + five)

prints 9, as istr are treated as ints (if possible).

Please note that four could have also been initialized with

four = istr(4)

or even

four, five = istr(4, 5)

Important

All calculations are strictly integer calculations. That means that if a float or decimal variable is ever produced it will be converted to an int. Also divisions are always floor divisions!

Use istr as str

We should realize that istrs are in fact strings.

In order to concatenate two istrs (or an istr and a str), we cannot use the + operator (remember four + five is istr('9')).

In order to concatenate istrs, we use the or operator (|). So

four | five

will be istr(45`).

And

(four | five) / 3

is istr('9').

In order to repeat a string in the usual sense, you cannot use the * operator (remember 3 * four is istr('12').

In order to repeat we use the matrix multiplication operator (@). So

3 @ four

is istr('444')

And

four @ 3

is also istr('444')

Note

It is not allowed to use the @ operator for two istrs. So, four @ five raises a TypeError.

istr that can't be interpreted as an int

Although usualy istrs are to be interpreted as an int, that's not a requirement.

So

istr('abc')

or

istr('1,2,3')

are perfectly acceptable.

But, we can't do any arithmetic or comparison with them.

If we try

istr('abc') + 5

a TypeError will be raised.

That holds for any arithmetic we try.

If we want to test if an istr can be interpreted (and thus used in an arithmetic and comparison expression). we can use the is_int() method. So

ìstr(20).is_int()

is True, whereas

ìstr('abc').is_int()

is False.

The bool operator works normally on the integer value of an istr. So

bool(istr('0')) ==> False bool(istr('1')) ==> True

But if the istr can't be interpreted as an int, the string value will be used to test. So

bool(istr('abc')) ==> True bool(istr('')) ==> False

Other operators

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'

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 values that are neither string nor numeric to initialize istr

Apart from with 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({'one': 1, 'two':2}) ==> {'one': istr('1'), 'two': istr('2')}
  

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

  mapped = (i for i in istr((i for i in range(2))))
  print(mapped)
  print(list(mapped))
  

  this wil print something like

  <generator object <genexpr> at 0x000002A10DE569B0>
  [istr('0'), istr('1')]
  

• 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

  istr(istr.range(5)) ==> istr.range(5)
  

• if an istr, the same istr will be returned

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

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 (provided the istr can be interpreted as an int) with the

is_even and is_odd method, e.g.

istr(4).is_even()) ==> True
istr(5).is_odd()) ==> True

test whether all characters are distinct

With the all_distinct method, it is possible to test whether all characters are distinct (i.e. no character appearts more than once).

istr('01234').all_distict() ==> True
istr('012340').all_distict() ==> False
n98 = istr(98)
n100 = n98 + 2
istr(n98).all_distinct() ==> True
istr(n100).all_distinct() ==> False

reverse an istr

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

istr(456).reversed() ==> istr('654')
istr('0456').reversed() ==> istr('6540')

The same can -of course- be achieved with

istr(456)[::-1] ==> istr('654')
istr('0456')[::-1] ==> istr('6540')

Note

It is possible to reverse a negative istr, but the result can't be interpreted as an int anymore.

istr(-456).reversed() + 3 ==> TypeError

enumerate with istrs

The istr.enumerate class method can be used just as the built-in 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.concat 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')] 

generate istr with digits

The class method digits can be used to return an istr of digits according to a given specification. The method takes either no or a number of arguments.

If no arguments are given, the result will be istr('0123456789').

The given argument(s) result in a range of digits.

• <n> ==> n
• <n-m> ==> n, n+1, ..., m
• -n> ==> 0, 1, ... n
• n-> ==> n, n+1, ..., 9 if n is numeric (0-9), n, n+1, ... Z if n is a letter
• '-' ==> 0, 1, ..., 9
• '' ==> 0, 1, ..., 9

(n and m must be digits between 0 and 9 or letters letters between A and Z)

When no stop value is specified, it will be

• 9 if the start value is between 0 and 9
• Z if the start value is between A and Z

The final result is an istr composed of the given range(s).

Here are some examples:

istr.digits() ==> istr('0123456789')
istr.digits('') ==> istr('0123456789')
istr.digits('1') ==> istr('1')
istr.digits('3-') ==> istr('3456789')
istr.digits('-3') ==> istr('0123')
istr('1-4', '6', '8-9') ==> istr('1234689')
istr('1', '1-2', '1-3') ==> istr('11213')
istr.digits('-z') ==> istr('0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ')
istr.digits('A-F') ==> istr('ABCDEF')
istr.digits('C') ==> istr('C')
istr.digits('3-') ==> istr('34567879')
istr.digits('X-') ==> istr('XYZ')

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, 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')

If the repr_mode is 'int' and the istr can't be interpreted as an int the string ? will be returned:

 with istr.repr_mode('int'):
    abc = istr('abc')
    print(repr(abc))

This will print

?

Note

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(127)
    print(repr(b))

This will result in

32767
istr('7F')

All calculations are done in the decimal 10 base 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

When an istr is initialized with a string the istr will be always stored as such.

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

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

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

With the istr.int_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.int_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.int_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

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

Overview of operations

The table below shows whether the string or the int version of istr is applied.

operator/function   int  str   Example
-----------------------------------------------------------------------------------------
+                    x         istr(20) + 3 ==> istr('23')
_                    x         istr(20) - 3 ==> istr('17')
*                    x         istr(20) * 3 ==> istr('60')
/                    x         istr(20) / 3 ==> istr('6')
//                   x         istr(20) // 3 ==> istr('6')
%                    x         istr(20) % 3 ==> istr('2')
divmod               x         divmod(istr(20), 3) ==> (istr('6'), istr('2'))
**                   x         istr(2) ** 3 ==> istr('8')
@                         x    istr(20) @ 3 ==> istr('202020')
==                   x    x    istr(20) == 20 ==> True | istr(20) == '20' ==> True
|                         x    istr(20) | 5 ==> istr('205')
abs                  x         abs(istr(-20)) ==> istr('20')
bool                 x    x *) bool(istr(' 0 ')) ==> False | istr('') ==> False
<=, <, >, >=         x         istr('100') > istr('2') ==> True
slicing                   x    istr(12345)[1:3] ==> istr('23')
iterate                   x    [x for x in istr(20)] ==> [istr('2'), istr('0')
len                       x    len(istr(' 20 ')) ==> 4
count                     x    istr(100),count('0') ==> 2
index                     x    istr(' 100 ').index('0') ==> 2
split                     x    istr('1 2').split() ==> (istr('1'), istr('2'))
string format             x    f"|{istr((1234):6}|" ==> '|1234  |'
other string methods      x    istr('aAbBcC').lower() ==> istr('aabbcc')
                               istr('aAbBcC').islower() ==> False
                               istr('  abc   ').strip() ==> istr('abc')
-----------------------------------------------------------------------------------------
*) str is applied if is_int() is False

Test script

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

This script also shows clearly the ways istr can be used, including several edge cases. Highly recommended to have a look at.

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