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.
Related
I have some number 0.0000002345E^-60. I want to print the floating point value as it is.
What is the way to do it?
print %f truncates it to 6 digits. Also %n.nf gives fixed numbers. What is the way to print without truncation.
Like this?
>>> print('{:.100f}'.format(0.0000002345E-60))
0.0000000000000000000000000000000000000000000000000000000000000000002344999999999999860343602938602754
As you might notice from the output, it’s not really that clear how you want to do it. Due to the float representation you lose precision and can’t really represent the number precisely. As such it’s not really clear where you want the number to stop displaying.
Also note that the exponential representation is often used to more explicitly show the number of significant digits the number has.
You could also use decimal to not lose the precision due to binary float truncation:
>>> from decimal import Decimal
>>> d = Decimal('0.0000002345E-60')
>>> p = abs(d.as_tuple().exponent)
>>> print(('{:.%df}' % p).format(d))
0.0000000000000000000000000000000000000000000000000000000000000000002345
You can use decimal.Decimal:
>>> from decimal import Decimal
>>> str(Decimal(0.0000002345e-60))
'2.344999999999999860343602938602754401109865640550232148836753621775217856801120686600683401464097113374472942165409862789978024748827516129306833728589548440037314681709534891496105046826414763927459716796875E-67'
This is the actual value of float created by literal 0.0000002345e-60. Its value is a number representable as python float which is closest to actual 0.0000002345 * 10**-60.
float should be generally used for approximate calculations. If you want accurate results you should use something else, like mentioned Decimal.
If I understand, you want to print a float?
The problem is, you cannot print a float.
You can only print a string representation of a float. So, in short, you cannot print a float, that is your answer.
If you accept that you need to print a string representation of a float, and your question is how specify your preferred format for the string representations of your floats, then judging by the comments you have been very unclear in your question.
If you would like to print the string representations of your floats in exponent notation, then the format specification language allows this:
{:g} or {:G}, depending whether or not you want the E in the output to be capitalized). This gets around the default precision for e and E types, which leads to unwanted trailing 0s in the part before the exponent symbol.
Assuming your value is my_float, "{:G}".format(my_float) would print the output the way that the Python interpreter prints it. You could probably just print the number without any formatting and get the same exact result.
If your goal is to print the string representation of the float with its current precision, in non-exponentiated form, User poke describes a good way to do this by casting the float to a Decimal object.
If, for some reason, you do not want to do this, you can do something like is mentioned in this answer. However, you should set 'max_digits' to sys.float_info.max_10_exp, instead of 14 used in the answer. This requires you to import sys at some point prior in the code.
A full example of this would be:
import math
import sys
def precision_and_scale(x):
max_digits = sys.float_info.max_10_exp
int_part = int(abs(x))
magnitude = 1 if int_part == 0 else int(math.log10(int_part)) + 1
if magnitude >= max_digits:
return (magnitude, 0)
frac_part = abs(x) - int_part
multiplier = 10 ** (max_digits - magnitude)
frac_digits = multiplier + int(multiplier * frac_part + 0.5)
while frac_digits % 10 == 0:
frac_digits /= 10
scale = int(math.log10(frac_digits))
return (magnitude + scale, scale)
f = 0.0000002345E^-60
p, s = precision_and_scale(f)
print "{:.{p}f}".format(f, p=p)
But I think the method involving casting to Decimal is probably better, overall.
I need to record SerialNumber(s) on an object. We enter many objects. Most serial numbers are strings - the numbers aren't used numerically, just as unique identifiers - but they are often sequential. Further, leading zeros are important due to unique id status of serial number.
When doing data entry, it's nice to just enter the first "sequential" serial number (eg 000123) and then the number of items (eg 5) to get the desired output - that way we can enter data in bulk see below:
Obj1.serial = 000123
Obj2.serial = 000124
Obj3.serial = 000125
Obj4.serial = 000126
Obj5.serial = 000127
The problem is that when you take the first number-as-string, turn to integer and increment, you loose the leading zeros.
Not all serials are sequential - not all are even numbers (eg FDM-434\RRTASDVI908)
But those that are, I would like to automate entry.
In python, what is the most elegant way to check for leading zeros (*and, I guess, edge cases like 0009999) in a string before iterating, and then re-application of those zeros after increment?
I have a solution to this problem but it isn't elegant. In fact, it's the most boring and blunt alg possible.
Is there an elegant solution to this problem?
EDIT
To clarify the question, I want the serial to have the same number of digits after the increment.
So, in most cases, this will mean reapplying the same number of leading zeros. BUT in some edge cases the number of leading zeros will be decremented. eg: 009 -> 010; 0099 -> 0100
Try str.zfill():
>>> s = "000123"
>>> i = int(s)
>>> i
123
>>> n = 6
>>> str(i).zfill(n)
'000123'
I develop my comment here, Obj1.serial being a string:
Obj1.serial = "000123"
('%0'+str(len(Obj1.serial))+'d') % (1+int(Obj1.serial))
It's like #owen-s answer '%06d' % n: print the number and pad with leading 0.
Regarding '%d' % n, it's just one way of printing. From PEP3101:
In Python 3.0, the % operator is supplemented by a more powerful
string formatting method, format(). Support for the str.format()
method has been backported to Python 2.6.
So you may want to use format instead… Anyway, you have an integer at the right of the % sign, and it will replace the %d inside the left string.
'%06d' means print a minimum of 6 (6) digits (d) long, fill with 0 (0) if necessary.
As Obj1.serial is a string, you have to convert it to an integer before the increment: 1+int(Obj1.serial). And because the right side takes an integer, we can leave it like that.
Now, for the left part, as we can't hard code 6, we have to take the length of Obj1.serial. But this is an integer, so we have to convert it back to a string, and concatenate to the rest of the expression %0 6 d : '%0'+str(len(Obj1.serial))+'d'. Thus
('%0'+str(len(Obj1.serial))+'d') % (1+int(Obj1.serial))
Now, with format (format-specification):
'{0:06}'.format(n)
is replaced in the same way by
('{0:0'+str(len(Obj1.serial))+'}').format(1+int(Obj1.serial))
You could check the length of the string ahead of time, then use rjust to pad to the same length afterwards:
>>> s = "000123"
>>> len_s = len(s)
>>> i = int(s)
>>> i
123
>>> str(i).rjust(len_s, "0")
'000123'
You can check a serial number for all digits using:
if serial.isdigit():
Let us suppose that we have a string representing a binary fraction such as:
".1"
As a decimal number this is 0.5. Is there a standard way in Python to go from such strings to a number type (whether it is binary or decimal is not strictly important).
For an integer, the solution is straightforward:
int("101", 2)
>>>5
int() takes an optional second argument to provide the base, but float() does not.
I am looking for something functionally equivalent (I think) to this:
def frac_bin_str_to_float(num):
"""Assuming num to be a string representing
the fractional part of a binary number with
no integer part, return num as a float."""
result = 0
ex = 2.0
for c in num:
if c == '1':
result += 1/ex
ex *= 2
return result
I think that does what I want, although I may well have missed some edge cases.
Is there a built-in or standard method of doing this in Python?
The following is a shorter way to express the same algorithm:
def parse_bin(s):
return int(s[1:], 2) / 2.**(len(s) - 1)
It assumes that the string starts with the dot. If you want something more general, the following will handle both the integer and the fractional parts:
def parse_bin(s):
t = s.split('.')
return int(t[0], 2) + int(t[1], 2) / 2.**len(t[1])
For example:
In [56]: parse_bin('10.11')
Out[56]: 2.75
It is reasonable to suppress the point instead of splitting on it, as follows. This bin2float function (unlike parse_bin in previous answer) correctly deals with inputs without points (except for returning an integer instead of a float in that case).
For example, the invocations bin2float('101101'), bin2float('.11101'), andbin2float('101101.11101')` return 45, 0.90625, 45.90625 respectively.
def bin2float (b):
s, f = b.find('.')+1, int(b.replace('.',''), 2)
return f/2.**(len(b)-s) if s else f
You could actually generalize James's code to convert it from any number system if you replace the hard coded '2' to that base.
def str2float(s, base=10):
dot, f = s.find('.') + 1, int(s.replace('.', ''), base)
return f / float(base)**(len(s) - dot) if dot else f
You can use the Binary fractions package. With this package you can convert binary-fraction strings into floats and vice-versa.
Example:
>>> from binary_fractions import Binary
>>> float(Binary("0.1"))
0.5
>>> str(Binary(0.5))
'0b0.1'
It has many more helper functions to manipulate binary strings such as: shift, add, fill, to_exponential, invert...
PS: Shameless plug, I'm the author of this package.
How can I get the x LSBs from a string (str) in Python?
In the specific I have a 256 bits string consisting in 32 chars each occupying 1 byte, from which I have to get a "char" string with the 50 Least Significant Bits.
So here are the ingredients for an approach that works using strings (simple but not the most efficient variant):
ord(x) yields the number (i.e. essentially the bits) for a char (e.g. ord('A')=65). Note that ord expects really an byte-long character (no special signs such as € or similar...)
bin(x)[2:] creates a string representing the number x in binary.
Thus, we can do (mystr holds your string):
l = [bin(ord(x))[2:] for x in mystr] # retrieve every character as binary number
bits = ""
for x in l: # concatenate all bits
bits = bits + l
bits[-50:] # retrieve the last 50 bits
Note that this approach is surely not the most efficient one due to the heavy string operations that could be replaced by plain integer operations (using bit-shifts and such). However, it is the simplest variant that came to my mind.
I think that a possible answer could be in this function:
mystr holds my string
def get_lsbs_str(mystr):
chrlist = list(mystr)
result1 = [chr(ord(chrlist[-7])&(3))]
result2 = chrlist[-6:]
return "".join(result1 + result2)
this function take 2 LSBs of the -7rd char of mystr (these are the 2 MSBs of the 50 LSBs)
then take the last 6 characters of mystr (these are the 48 LSB of the 50 LSB)
Please make me know if I am in error.
If it's only to display, wouldn't it help you?
yourString [14:63]
You can also use
yourString [-50:]
For more information, see here
I want to format a number with a decimal point in it with leading zeros.
This
>>> '3.3'.zfill(5)
003.3
considers all the digits and even the decimal point. Is there a function in python that considers only the whole part?
I only need to format simple numbers with no more than five decimal places. Also, using %5f seems to consider trailing instead of leading zeros.
Is that what you look for?
>>> "%07.1f" % 2.11
'00002.1'
So according to your comment, I can come up with this one (although not as elegant anymore):
>>> fmt = lambda x : "%04d" % x + str(x%1)[1:]
>>> fmt(3.1)
0003.1
>>> fmt(3.158)
0003.158
I like the new style of formatting.
loop = 2
pause = 2
print 'Begin Loop {0}, {1:06.2f} Seconds Pause'.format(loop, pause)
>>>Begin Loop 2, 0002.1 Seconds Pause
In {1:06.2f}:
1 is the place holder for variable pause
0 indicates to pad with leading zeros
6 total number of characters including the decimal point
2 the precision
f converts integers to floats
print('{0:07.3f}'.format(12.34))
This will have total 7 characters including 3 decimal points, ie. "012.340"
Like this?
>>> '%#05.1f' % 3.3
'003.3'
Starting with a string as your example does, you could write a small function such as this to do what you want:
def zpad(val, n):
bits = val.split('.')
return "%s.%s" % (bits[0].zfill(n), bits[1])
>>> zpad('3.3', 5)
'00003.3'
With Python 3.6+ you can use the fstring method:
f'{3.3:.0f}'[-5:]
>>> '3'
f'{30000.3:.0f}'[-5:]
>>> '30000'
This method will eliminate the fractional component (consider only the whole part) and return up to 5 digits. Two caveats: First, if the whole part is larger than 5 digits, the most significant digits beyond 5 will be removed.
Second, if the fractional component is greater than 0.5, the function will round up.
f'{300000.51:.0f}'[-5:]
>>>'00001'