Develop Reference

How can I densely store large numbers in a file?

I need to store and handle huge amounts of very long numbers, which are in range from 0 to f 64 times (ffffffffff.....ffff).
If I store these numbers in a file, I need 1 byte for each character (digit) + 2 bytes for \n symbol = up to 66 bytes. However to represent all possible numbers we need not more than 34 bytes (4 bits represent digits from 0 to f, therefore 4 [bits] * 64 [amount of hex digits]/8 [bits a in byte] = 32 bytes + \n, of course).
Is there any way to store the number without consuming excess memory?
So far I have created converter from hex (with 16 digits per symbol) to a number with base of 76 (hex + all letters and some other symbols), which reduces size of a number to 41 + 2 bytes.

You are trying to store 32 bytes long. Why not just store them as binary numbers? That way you need to store only 32 bytes per number instead of 41 or whatever. You can add on all sorts of quasi-compression schemes to take advantage of things like most of your numbers being shorter than 32 bytes.
If your number is a string, convert it to an int first. Python3 ints are basically infinite precision, so you will not lose any information:
>>> num = '113AB87C877AAE3790'
>>> num = int(num, 16)
>>> num
317825918024297625488
Now you can convert the result to a byte array and write it to a file opened for binary writing:
with open('output.bin', 'wb') as file:
file.write(num.to_bytes(32, byteorder='big'))
The int method to_bytes converts your number to a string of bytes that can be placed in a file. You need to specify the string length and the order. 'big' makes it easier to read a hex dump of the file.
To read the file back and decode it using int.from_bytes in a similar manner:
with open('output.bin', 'rb') as file:
bytes = file.read(32)
num = int.from_bytes(bytes, byteorder='big')
Remember to always include the b in the file mode, or you may run into unexpected problems if you try to read or write data with codes for \n in it.
Both the read and write operation can be looped as a matter of course.

If you anticipate storing an even distribution of numbers, then see Mad Physicist's answer. However, If you anticipate storing mostly small numbers but need to be able to store a few large numbers, then these schemes may also be useful.
If you only need to account for integers that are 255 or fewer bytes (2040 or fewer bits) in length, then simply convert the int to a bytes object and store the length in an additional byte, like this:
# This was only tested with non-negative integers!
def encode(num):
assert isinstance(num, int)
# Convert the number to a byte array and strip away leading null bytes.
# You can also use byteorder="little" and rstrip.
# If the integer does not fit into 255 bytes, an OverflowError will be raised.
encoded = num.to_bytes(255, byteorder="big").lstrip(b'\0')
# Return the length of the integer in the first byte, followed by the encoded integer.
return bytes([len(encoded)]) + encoded
def encode_many(nums):
return b''.join(encode(num) for num in nums)
def decode_many(byte_array):
assert isinstance(byte_array, bytes)
result = []
start = 0
while start < len(byte_array):
# The first byte contains the length of the integer.
int_length = byte_array[start]
# Read int_length bytes and decode them as int.
new_int = int.from_bytes(byte_array[(start+1):(start+int_length+1)], byteorder="big")
# Add the new integer to the result list.
result.append(new_int)
start += int_length + 1
return result
To store integers of (practically) infinite length, you can use this scheme, based on variable-length quantities in the MIDI file format. First, the rules:
A byte has eight bits (for those who don't know).
In each byte except the last, the left-most bit (the highest-order bit) will be 1.
The lower seven bits (i.e. all bits except the left-most bit) in each byte, when concatenated together, form an integer with a variable number of bits.
Here are a few examples:
0 in binary is 00000000. It can be represented in one byte without modification as 00000000.
127 in binary is 01111111. It can be represented in one byte without modification as 01111111.
128 in binary is 10000000. It must be converted to a two-byte representation: 10000001 00000000. Let's break that down:
The left-most bit in the first byte is 1, which means that it is not the last byte.
The left-most bit in the second byte is 0, which means that it is the last byte.
The lower seven bits in the first byte are 0000001, and the lower seven bits in the second byte are 0000000. Concatenate those together, and you get 00000010000000, which is 128.
173249806138790 in binary is 100111011001000111011101001001101111110110100110.
To store it:
First, split the binary number into groups of seven bits: 0100111 0110010 0011101 1101001 0011011 1111011 0100110 (a leading 0 was added)
Then, add a 1 in front of each byte except the last, which gets a 0: 10100111 10110010 10011101 11101001 10011011 11111011 00100110
To retrieve it:
First, drop the first bit of each byte: 0100111 0110010 0011101 1101001 0011011 1111011 0100110
You are left with an array of seven-bit segments. Join them together: 100111011001000111011101001001101111110110100110
When that is converted to decimal, you get 173,249,806,138,790.
Why, you ask, do we make the left-most bit in the last byte of each number a 0? Well, doing that allows you to concatenate multiple numbers together without using line breaks. When writing the numbers to a file, just write them one after another. When reading the numbers from a file, use a loop that builds an array of integers, ending each integer whenever it detects a byte where the left-most bit is 0.
Here are two functions, encode and decode, which convert between int and bytes in Python 3.
# Important! These methods only work with non-negative integers!
def encode(num):
assert isinstance(num, int)
# If the number is 0, then just return a single null byte.
if num <= 0:
return b'\0'
# Otherwise...
result_bytes_reversed = []
while num > 0:
# Find the right-most seven bits in the integer.
current_seven_bit_segment = num & 0b1111111
# Change the left-most bit to a 1.
current_seven_bit_segment |= 0b10000000
# Add that to the result array.
result_bytes_reversed.append(current_seven_bit_segment)
# Chop off the right-most seven bits.
num = num >> 7
# Change the left-most bit in the lowest-order byte (which is first in the list) back to a 0.
result_bytes_reversed[0] &= 0b1111111
# Un-reverse the order of the bytes and convert the list into a byte string.
return bytes(reversed(result_bytes_reversed))
def decode(byte_array):
assert isinstance(byte_array, bytes)
result = 0
for part in byte_array:
# Shift the result over by seven bits.
result = result << 7
# Add in the right-most seven bits from this part.
result |= (part & 0b1111111)
return result
Here are two functions for working with lists of ints:
def encode_many(nums):
return [encode(num) for num in nums]
def decode_many(byte_array):
parts = []
# Split the byte array after each byte where the left-most bit is 0.
start = 0
for i, b in enumerate(byte_array):
# Check whether the left-most bit in this byte is 0.
if not (b & 0b10000000):
# Copy everything up to here into a new part.
parts.append(byte_array[start:(i+1)])
start = i + 1
return [decode(part) for part in parts]

The densest possible way without knowing more about the numbers would be 256 bits per number (32 bytes).
You can store them right after one another.
A function to write to a file might look like this:
def write_numbers(numbers, file):
for n in numbers:
file.write(n.to_bytes(32, 'big'))
with open('file_name', 'wb') as f:
write_numbers(get_numbers(), f)
And to read the numbers, you can make a function like this:
def read_numbers(file):
while True:
read = file.read(32)
if not read:
break
yield int.from_bytes(read, 'big')
with open('file_name', 'rb') as f:
for n in read_numbers(f):
do_stuff(n)

Representing a word as sequence of bits

I want to represent a word as a sequence of 26 bits. If 25th bit is set it means that the letter 'y' is present in that word.
For example: word:"abekz"
representation:10000000000000010000010011
This is very easy to do it in C/C++ since it has a 32 bit int type. But Python's int has infinite precision so I'm unable to do it.
Here's my (Wrong)solution:
def representAsBits(string):
mask=0
for each_char in string:
bit_position= ord(each_char)-97 #string consists of only lower-case letters
mask= mask | (1<<bit_position)
return bin(mask)
print representAsBits("abze")# gives me 0b10000000000000000000010011
print representAsBits("wxcc")# gives me 0b110000000000000000000100 2 bits missing here
What changes can I make? Thanks!

You can't store leading zeroes on an integer. Thankfully, you're using bin(), which returns a string.
With a little creative slicing, we can format it however we want:
return "0b%32d" % int(bin(mask)[2:])
will give:
>>> representAsBits("abekz")
'0b00000010000000000000010000010011'
That being said, to compare masks, you don't have to bin() them except if you want to "show" the binary. Compare the integers themselves, which will be the same:
with return mask:
>>> representAsBits("z") == representAsBits("zzz")
True
Although, since the masks will match, it doesn't matter what padding you use, as they will be the same if generated from the same mask: Any string containing only the characters wxc will yield the same string, regardless of what method you use.

STL binary file reader with Python

I'm trying to write my "personal" python version of STL binary file reader, according to WIKIPEDIA : A binary STL file contains :
an 80-character (byte) headern which is generally ignored.
a 4-byte unsigned integer indicating the number of triangular facets in the file.
Each triangle is described by twelve 32-bit floating-point numbers: three for the normal and then three for the X/Y/Z coordinate of each vertex – just as with the ASCII version of STL. After these follows a 2-byte ("short") unsigned integer that is the "attribute byte count" – in the standard format, this should be zero because most software does not understand anything else. --Floating-point numbers are represented as IEEE floating-point numbers and are assumed to be little-endian--
Here is my code :
#! /usr/bin/env python3
with open("stlbinaryfile.stl","rb") as fichier :
head=fichier.read(80)
nbtriangles=fichier.read(4)
print(nbtriangles)
The output is :
b'\x90\x08\x00\x00'
It represents an unsigned integer, I need to convert it without using any package (struct,stl...). Are there any (basic) rules to do it ?, I don't know what does \x mean ? How does \x90 represent one byte ?
most of the answers in google mention "C structs", but I don't know nothing about C.
Thank you for your time.

Since you're using Python 3, you can use int.from_bytes. I'm guessing the value is stored little-endian, so you'd just do:
nbtriangles = int.from_bytes(fichier.read(4), 'little')
Change the second argument to 'big' if it's supposed to be big-endian.
Mind you, the normal way to parse a fixed width type is the struct module, but apparently you've ruled that out.
For the confusion over the repr, bytes objects will display ASCII printable characters (e.g. a) or standard ASCII escapes (e.g. \t) if the byte value corresponds to one of them. If it doesn't, it uses \x##, where ## is the hexadecimal representation of the byte value, so \x90 represents the byte with value 0x90, or 144. You need to combine the byte values at offsets to reconstruct the int, but int.from_bytes does this for you faster than any hand-rolled solution could.
Update: Since apparent int.from_bytes isn't "basic" enough, a couple more complex, but only using top-level built-ins (not alternate constructors) solutions. For little-endian, you can do this:
def int_from_bytes(inbytes):
res = 0
for i, b in enumerate(inbytes):
res |= b << (i * 8) # Adjust each byte individually by 8 times position
return res
You can use the same solution for big-endian by adding reversed to the loop, making it enumerate(reversed(inbytes)), or you can use this alternative solution that handles the offset adjustment a different way:
def int_from_bytes(inbytes):
res = 0
for b in inbytes:
res <<= 8 # Adjust bytes seen so far to make room for new byte
res |= b # Mask in new byte
return res
Again, this big-endian solution can trivially work for little-endian by looping over reversed(inbytes) instead of inbytes. In both cases inbytes[::-1] is an alternative to reversed(inbytes) (the former makes a new bytes in reversed order and iterates that, the latter iterates the existing bytes object in reverse, but unless it's a huge bytes object, enough to strain RAM if you copy it, the difference is pretty minimal).

The typical way to interpret an integer is to use struct.unpack, like so:
import struct
with open("stlbinaryfile.stl","rb") as fichier :
head=fichier.read(80)
nbtriangles=fichier.read(4)
print(nbtriangles)
nbtriangles=struct.unpack("<I", nbtriangles)
print(nbtriangles)
If you are allergic to import struct, then you can also compute it by hand:
def unsigned_int(s):
result = 0
for ch in s[::-1]:
result *= 256
result += ch
return result
...
nbtriangles = unsigned_int(nbtriangles)
As to what you are seeing when you print b'\x90\x08\x00\x00'. You are printing a bytes object, which is an array of integers in the range [0-255]. The first integer has the value 144 (decimal) or 90 (hexadecimal). When printing a bytes object, that value is represented by the string \x90. The 2nd has the value eight, represented by \x08. The 3rd and final integers are both zero. They are presented by \x00.
If you would like to see a more familiar representation of the integers, try:
print(list(nbtriangles))
[144, 8, 0, 0]
To compute the 32-bit integers represented by these four 8-bit integers, you can use this formula:
total = byte0 + (byte1*256) + (byte2*256*256) + (byte3*256*256*256)
Or, in hex:
total = byte0 + (byte1*0x100) + (byte2*0x10000) + (byte3*0x1000000)
Which results in:
0x00000890
Perhaps you can see the similarities to decimal, where the string "1234" represents the number:
4 + 3*10 + 2*100 + 1*1000

How to compute a double precision float score from the first 8 bytes of a string in Python?

Trying to get a double-precision floating point score from a UTF-8 encoded string object in Python. The idea is to grab the first 8 bytes of the string and create a float, so that the strings, ordered by their score, would be ordered lexicographically according to their first 8 bytes (or possibly their first 63 bits, after forcing them all to be positive to avoid sign errors).
For example:
get_score(u'aaaaaaa') < get_score(u'aaaaaaab') < get_score(u'zzzzzzzz')
I have tried to compute the score in an integer using bit-shift-left and XOR, but I am not sure of how to translate that into a float value. I am also not sure if there is a better way to do this.
How should the score for a string be computed so the condition I specified before is met?
Edit: The string object is UTF-8 encoded (as per #Bakuriu's commment).

float won't give you 64 bits of precision. Use integers instead.
def get_score(s):
return struct.unpack('>Q', (u'\0\0\0\0\0\0\0\0' + s[:8])[-8:])[0]
In Python 3:
def get_score(s):
return struct.unpack('>Q', ('\0\0\0\0\0\0\0\0' + s[:8])[-8:].encode('ascii', 'error'))[0]
EDIT:
For floats, with 6 characters:
def get_score(s):
return struct.unpack('>d', (u'\0\1' + (u'\0\0\0\0\0\0\0\0' + s[:6])[-6:]).encode('ascii', 'error'))[0]

You will need to setup the entire alphabet and do the conversion by hand, since conversions to base > 36 are not built in, in order to do that you only need to define the complete alphabet to use. If it was an ascii string for instance you would create a conversion to a long in base 256 from the input string using all the ascii table as an alphabet.
You have an example of the full functions to do it here: string to base 62 number
Also you don't need to worry about negative-positive numbers when doing this, since the encoding of the string with the first character in the alphabet will yield the minimum possible number in the representation, which is the negative value with the highest absolute value, in your case -2**63 which is the correct value and allows you to use < > against it.
Hope it helps!

Length of hexadecimal number

How can we get the length of a hexadecimal number in the Python language?
I tried using this code but even this is showing some error.
i = 0
def hex_len(a):
if a > 0x0:
# i = 0
i = i + 1
a = a/16
return i
b = 0x346
print(hex_len(b))
Here I just used 346 as the hexadecimal number, but my actual numbers are very big to be counted manually.

Use the function hex:
>>> b = 0x346
>>> hex(b)
'0x346'
>>> len(hex(b))-2
3
or using string formatting:
>>> len("{:x}".format(b))
3

While using the string representation as intermediate result has some merits in simplicity it's somewhat wasted time and memory. I'd prefer a mathematical solution (returning the pure number of digits without any 0x-prefix):
from math import ceil, log
def numberLength(n, base=16):
return ceil(log(n+1)/log(base))
The +1 adjustment takes care of the fact, that for an exact power of your number base you need a leading "1".

As Ashwini wrote, the hex function does the hard work for you:
hex(x)
Convert an integer number (of any size) to a hexadecimal string. The result is a valid Python expression.