For a project in one of my classes we have to output numbers up to five decimal places.It is possible that the output will be a complex number and I am unable to figure out how to output a complex number with five decimal places. For floats I know it is just:
print "%0.5f"%variable_name
Is there something similar for complex numbers?
You could do it as is shown below using the str.format() method:
>>> n = 3.4+2.3j
>>> n
(3.4+2.3j)
>>> '({0.real:.2f} + {0.imag:.2f}i)'.format(n)
'(3.40 + 2.30i)'
>>> '({c.real:.2f} + {c.imag:.2f}i)'.format(c=n)
'(3.40 + 2.30i)'
To make it handle both positive and negative imaginary portions properly, you would need a (even more) complicated formatting operation:
>>> n = 3.4-2.3j
>>> n
(3.4-2.3j)
>>> '({0:.2f} {1} {2:.2f}i)'.format(n.real, '+-'[n.imag < 0], abs(n.imag))
'(3.40 - 2.30i)'
Update - Easier Way
Although you cannot use f as a presentation type for complex numbers using the string formatting operator %:
n1 = 3.4+2.3j
n2 = 3.4-2.3j
try:
print('test: %.2f' % n1)
except Exception as exc:
print('{}: {}'.format(type(exc).__name__, exc))
Output:
TypeError: float argument required, not complex
You can however use it with complex numbers via the str.format() method. This isn't explicitly documented, but is implied by the Format Specification Mini-Language documentation which just says:
'f'Â Â Fixed point. Displays the number as a fixed-point number. The default precision is 6.
. . .so it's easy to overlook.
In concrete terms, the following works in both Python 2.7.14 and 3.4.6:
print('n1: {:.2f}'.format(n1))
print('n2: {:.2f}'.format(n2))
Output:
n1: 3.10+4.20j
n2: 3.10-4.20j
This doesn't give you quite the control the code in my original answer does, but it's certainly much more concise (and handles both positive and negative imaginary parts automatically).
Update 2 - f-strings
Formatted string literals (aka f-strings) were added in Python 3.6, which means it could also be done like this in that version or later:
print(f'n1: {n1:.2f}') # -> n1: 3.40+2.30j
print(f'n2: {n2:.3f}') # -> n2: 3.400-2.300j
In Python 3.8.0, support for an = specifier was added to f-strings, allowing you to write:
print(f'{n1=:.2f}') # -> n1=3.40+2.30j
print(f'{n2=:.3f}') # -> n2=3.400-2.300j
Neither String Formatting Operations - i.e. the modulo (%) operator) -
nor the newer str.format() Format String Syntax support complex types.
However it is possible to call the __format__ method of all built in numeric types directly.
Here is an example:
>>> i = -3 # int
>>> l = -33L # long (only Python 2.X)
>>> f = -10./3 # float
>>> c = - 1./9 - 2.j/9 # complex
>>> [ x.__format__('.3f') for x in (i, l, f, c)]
['-3.000', '-33.000', '-3.333', '-0.111-0.222j']
Note, that this works well with negative imaginary parts too.
For questions like this, the Python documentation should be your first stop. Specifically, have a look at the section on string formatting. It lists all the string format codes; there isn't one for complex numbers.
What you can do is format the real and imaginary parts of the number separately, using x.real and x.imag, and print it out in a + bi form.
>>> n = 3.4 + 2.3j
>>> print '%05f %05fi' % (n.real, n.imag)
3.400000 2.300000i
As of Python 2.6 you can define how objects of your own classes respond to format strings. So, you can define a subclass of complex that can be formatted. Here's an example:
>>> class Complex_formatted(complex):
... def __format__(self, fmt):
... cfmt = "({:" + fmt + "}{:+" + fmt + "}j)"
... return cfmt.format(self.real, self.imag)
...
>>> z1 = Complex_formatted(.123456789 + 123.456789j)
>>> z2 = Complex_formatted(.123456789 - 123.456789j)
>>> "My complex numbers are {:0.5f} and {:0.5f}.".format(z1, z2)
'My complex numbers are (0.12346+123.45679j) and (0.12346-123.45679j).'
>>> "My complex numbers are {:0.6f} and {:0.6f}.".format(z1, z2)
'My complex numbers are (0.123457+123.456789j) and (0.123457-123.456789j).'
Objects of this class behave exactly like complex numbers except they take more space and operate more slowly; reader beware.
Check this out:
np.set_printoptions(precision=2) # Rounds up to 2 decimals all float expressions
I've successfully printed my complexfloat's expressions:
# Show poles and zeros
print( "zeros = ", zeros_H , "\n")
print( "poles = ", poles_H )
out before:
zeros = [-0.8 +0.6j -0.8 -0.6j -0.66666667+0.j ]
poles = [-0.81542318+0.60991027j -0.81542318-0.60991027j -0.8358203 +0.j ]
out after:
zeros = [-0.8 +0.6j -0.8 -0.6j -0.67+0.j ]
poles = [-0.82+0.61j -0.82-0.61j -0.84+0.j ]
Related
I'd like to be able to decode an emoji into its corresponding code points as seen here. I'm limited to using standard libraries in 2.7.
For example:
🇲🇩 -> U+1F1F2 U+1F1E9
I've managed to get the first code point using this code, but I can't figure out how to pull the second. Some emoji have even more code points.
to_decode = u'🇲🇩'
code = ord(to_decode[0])
if 0xd800 <= code <= 0xdbff:
code = (code - 0xd800) * 1024 + (ord(to_decode[1]) - 0xdc00) + + 0x010000
print(hex(code))
A combination of encode and struct.unpack can give you what you need.
>>> import struct
>>> b = to_decode.encode('utf_32_le')
>>> count = len(b) // 4
>>> count
2
>>> cp = struct.unpack('<%dI' % count, b)
>>> [hex(x) for x in cp]
['0x1f1f2', '0x1f1e9']
This is sort of an hack, but you can use the repr of the unicode string:
>>> repr(to_decode)
"u'\\U0001f1f2\\U0001f1e9'"
so:
>>> hex(int(repr(to_decode)[4:12], 16))
'0x1f1f2'
and
>>> hex(int(repr(to_decode)[14:22], 16))
'0x1f1e9'
You must extend this method to support emojis with more than two code points. You may consider using a combination of the above with .split("\\U").
For this problem, you actually need list() which will break a Unicode character into its constituent code points
to_decode = u'🇲🇩'
list(to_decode)
['🇲', '🇩']
As an example of what you can do with this, I created a unicode visualization of the Bengali Alphabet
https://www.kaggle.com/jamesmcguigan/unicode-visualization-of-the-bengali-alphabet
I am trying to create a python code which can help to increment the version values below by 1,expected output shows the result?I am thinking of splitting the version saving each digit ,increment the last digit by 1 and reconstruct the version,is there a way to this simply in python?
version1 = 1151.1
version2 = 4.1.1
version3 = 275.1.2.99
version4 = 379
next_version1 = version1 + 1
print next_version1
next_version2 = version2 + 1
print next_version2
next_version3 = version3 + 1
print next_version3
next_version4 = version4 + 1
print next_version4
EXPECTED OUTPUT:-
1151.2
4.1.2
275.1.2.100
380
Actually not all the numbers are floats in this case.
You should treat it as strings and update the last element.
version1 = '275.1.2.3'
version2 = '279'
version3 = '275.2.3.10'
def updateVersion(version):
if '.' in version:
version = version.split('.')
version[-1] = str(int(version[-1]) + 1)
version = '.'.join(version)
else:
version = str(int(version)+1)
return version
updateVersion(version1)
updateVersion(version2)
Output:
275.1.2.4
280
275.2.3.11
First and foremost please read about Floating Point Arithmetic: Issues and Limitations
Maybe that was the reason why you ask, not clear.
However, I suggest to save each part as an integer, e.g.
main_version_number = 1151
minor_version_number = 1
sub_version_number = 0
You could maybe have a data structure with those fields (a Version class maybe?) with appropriate methods.
Do not rely on floating point arithmetic.
First off, the code you outline would most certainly give a syntax error.
A number of the form 2 is an integer; 2.2, a floating point; but a 2.2.2, meaningless.
You are looking for tuples here. For instance,
>>> version3 = (275,1,2,3)
Then you would get
>>> version3
(275, 1, 2, 3)
To dirty-update only the last bit of such a tuple, you could do
>>> version3 = version3[:-1] + (version3[-1] + 1,)
>>> version3
(275, 1, 2, 4)
The reason I call this dirty updating is that it will not take care of carrying over into the next significant bit.
Here's a relatively simple script to do just that that I could put together in a couple of minutes. Assuming you have stored your version number as a tuple object called version, attempt the following:
new_version = version
for bit in range(len(version)):
new_version = new_version[:-1-bit] + ((new_version[-1-bit] + 1) % 10,) + new_version[-bit:]
if -2-bit >=0:
new_version = new_version[-2-bit:] + (new_version[-2-bit] + (version[-2-bit] + 1) // 10,) + new_version[-1-bit:]
elif (version[-2-bit] + 1) // 10:
new_version = (1,) + new_version
Alternatively, take a look at bumpversion, a tool that lets you take care of version-numbering within your project, with git integration.
The variables 'version2' and 'version3' will result in a syntax error. This syntax error is caused by the fact that Python does not know of any (number) type that has several points in its value. In essence you are trying to use certain types in a way that they are not meant to be used. More specifically the floating point number is not suitable for your goals. As the name suggests a floating point number, only contains one point and that point can be placed anywhere between its digits (floating).
My advice would be to create your own type/class. This would enable you to store the version number in a way that allows for easy modification of its values and better separation of concerns in your code (i.e. that each part of your code is only concerned with one thing).
Example
class VersionNumber:
"""Represents a version number that can contain parts (major, minor, etc)."""
def __init__(self, *argv):
"""This is the constructor, i.e. a function that is called when you create a new VersionNumber.
The '*argv' allows the user of this class to give a variable amount of arguments. This is why
you can have a version number with only 1 number, and one with 4. The '*argv' is iterable."""
#Create a class specific field, that stores all the version number parts in a list.
self.Parts = []
#Fill it with the supplied arguments.
for part in argv:
self.Parts.append(part)
def __repr__(self):
"""This function is called when the VersionNumber needs to be displayed in the console"""
return str(self)
def __str__(self):
"""This function is called when the VersionNumber is parsed to a string"""
return '.'.join(map(str,self.Parts))
def incrementVersion(self, position, incrementAmount):
"""This function allows you to increment the version number. It does this by adjusting the list
we have set in the constructor."""
self.Parts[position] += incrementAmount
version1 = VersionNumber(1, 23, 45, 0)
print(version1)
#Position -1, takes the last (most right) version number part from the version number.
version1.incrementVersion(-1, 1)
print(version1)
version2 = VersionNumber(346)
print(version2)
version2.incrementVersion(-1, 2)
print(version2)
(update)
Here's the actual problem I'm seeing. Note that round() doesn't seem to be doing the trick.
Here's my code:
t0=time.time()
# stuff
t1=time.time()
perfdat={'et1' : round(t1-t0,6), 'et2': '%.6f'%(t1-t0)}
And the dict and json output, respectively:
{'et2': '0.010214', 'et1': 0.010214000000000001}
{"et2":"0.010214","et1":0.010214000000000001}
(end update)
I've got a floating point value that has a lot of extra digits of precision that I don't need. Is there a way to truncate those digits when formatting a json string?
I can get the truncation I need if I format the value as a string, but I would like to transmit the value as a (truncated) number.
import json
v=2.030000002
json.dumps({'x':v}) # would like to just have 2.030
'{"x": 2.030000002}'
s= '%.3f' % (v) # like this, but not as a string
json.dumps({'x' : s})
'{"x": "2.030"}'
Wrap the number into a float:
>>> s = float('%.3f' % (v))
>>> json.dumps({'x' : s})
{"x": 2.03}
Builtin function round can help
In [16]: v=2.030000002
In [17]: json.dumps({'x': round(v, 3)})
Out[17]: '{"x": 2.03}'
This is something I found from from the Python Standard library:
"Unlike hardware based binary floating point, the decimal module has a user alterable precision (defaulting to 28 places) which can be as large as needed for a given problem:
>>> from decimal import *
>>> getcontext().prec = 6
>>> Decimal(1) / Decimal(7)
Decimal('0.142857')
>>> getcontext().prec = 28
>>> Decimal(1) / Decimal(7)
Decimal('0.1428571428571428571428571429')
"
A better import statement would be:
from decimal import getcontext, Decimal
Then you could apply those same functions to specify an arbitrary precision. Hope this helps! I haven't actually used this before.
For your case: (still has the trailing zero issue)
getcontext().prec = 3
s = '2.030'
var = float(Decimal(s))
var returns 2.03
This following approach seems promising:
import json
v = 2.030000002
result = []
for part in json.JSONEncoder().iterencode({'x': v}):
try:
tmp = round(float(part), 3)
except ValueError:
pass
else:
part = '{:.3f}'.format(tmp)
result.append(part)
result = ''.join(result)
print result # -> {"x": 2.030}
Encountered a problem whereby my JSON data gets printed as a scientific notation instead of a float.
import urllib2
import json
import sys
url = 'https://bittrex.com/api/v1.1/public/getmarketsummary?market=btc-quid'
json_obj = urllib2.urlopen(url)
QUID_data = json.load(json_obj)
QUID_MarketName_Trex = QUID_data["result"][0]["MarketName"][4:9]
QUID_Last_Trex = QUID_data["result"][0]["Last"]
QUID_High_Trex = QUID_data["result"][0]["High"]
QUID_Low_Trex = QUID_data["result"][0]["Low"]
QUID_Volume_Trex = QUID_data["result"][0]["Volume"]
QUID_BaseVolume_Trex = QUID_data["result"][0]["BaseVolume"]
QUID_TimeStamp_Trex = QUID_data["result"][0]["TimeStamp"]
QUID_Bid_Trex = QUID_data["result"][0]["Bid"]
QUID_Ask_Trex = QUID_data["result"][0]["Ask"]
QUID_OpenBuyOrders_Trex = QUID_data["result"][0]["OpenBuyOrders"]
QUID_OpenSellOrders_Trex = QUID_data["result"][0]["OpenSellOrders"]
QUID_PrevDay_Trex = QUID_data["result"][0]["PrevDay"]
QUID_Created_Trex = QUID_data["result"][0]["Created"]
QUID_Change_Trex = ((QUID_Last_Trex - QUID_PrevDay_Trex)/ QUID_PrevDay_Trex)*100
QUID_Change_Var = str(QUID_Change_Trex)
QUID_Change_Final = QUID_Change_Var[0:5] + '%'
print QUID_Last_Trex
It prints the following value; 1.357e-05.
I need this to be a float with 8 chars behind the decimal (0.00001370)
As you can see here --> http://i.imgur.com/FCVM1UN.jpg, my GUI displays the first row correct (using the exact same code).
You are looking at the default str() formatting of floating point numbers, where scientific notation is used for sufficiently small or large numbers.
You don't need to convert this, the value itself is a proper float. If you need to display this in a different format, format it explicitly:
>>> print(0.00001357)
1.357e-05
>>> print(format(0.00001357, 'f'))
0.000014
>>> print(format(0.00001357, '.8f'))
0.00001357
Here the f format always uses fixed point notation for the value. The default precision is 6 digits; the .8 instructs the f formatter to show 8 digits instead.
In Python 3, the default string format is essentially the same as format(fpvalue, '.16g'); the g format uses either a scientific or fixed point presentation depending on the exponent of the number. Python 2 used '.12g'.
You can use print formatting:
x = 1.357e-05
print('%f' % x)
Edit:
print('%.08f' % x)
There are some approaches:
#1 float(...) + optionally round() or .format()
x = float(1.357e-05)
round(x, 6)
"{:.8f}".format(x)
#2 with decimal class
import decimal
tmp = decimal.Decimal('1.357e-05')
print('[0]', tmp)
# [0] 0.00001357
tmp = decimal.Decimal(1.357e-05)
print('[1]', tmp)
# [1] 0.0000135700000000000005188384444299032338676624931395053863525390625
decimal.getcontext().prec = 6
tmp = decimal.getcontext().create_decimal(1.357e-05)
print('[2]', tmp)
# [2] 0.0000135700
#3 with .rstrip(...)
x = ("%.17f" % n).rstrip('0').rstrip('.')
Note: there are counterparts to %f:
%f shows standard notation
%e shows scientific notation
%g shows default (scientific if 5 or more zeroes)
I need the Python analog for this Perl string:
unpack("nNccH*", string_val)
I need the nNccH* - data format in Python format characters.
In Perl it unpack binary data to five variables:
16 bit value in "network" (big-endian)
32 bit value in "network" (big-endian)
Signed char (8-bit integer) value
Signed char (8-bit integer) value
Hexadecimal string, high nibble first
But I can't do it in Python
More:
bstring = ''
while DataByte = client[0].recv(1):
bstring += DataByte
print len(bstring)
if len(bstring):
a, b, c, d, e = unpack("nNccH*", bstring)
I never wrote in Perl or Python, but my current task is to write a multithreading Python server that was written in Perl...
The Perl format "nNcc" is equivalent to the Python format "!HLbb".
There is no direct equivalent in Python for Perl's "H*".
There are two problems.
Python's struct.unpack does not accept the wildcard character, *
Python's struct.unpack does not "hexlify" data strings
The first problem can be worked-around using a helper function like unpack.
The second problem can be solved using binascii.hexlify:
import struct
import binascii
def unpack(fmt, data):
"""
Return struct.unpack(fmt, data) with the optional single * in fmt replaced with
the appropriate number, given the length of data.
"""
# http://stackoverflow.com/a/7867892/190597
try:
return struct.unpack(fmt, data)
except struct.error:
flen = struct.calcsize(fmt.replace('*', ''))
alen = len(data)
idx = fmt.find('*')
before_char = fmt[idx-1]
n = (alen-flen)//struct.calcsize(before_char)+1
fmt = ''.join((fmt[:idx-1], str(n), before_char, fmt[idx+1:]))
return struct.unpack(fmt, data)
data = open('data').read()
x = list(unpack("!HLbbs*", data))
# x[-1].encode('hex') works in Python 2, but not in Python 3
x[-1] = binascii.hexlify(x[-1])
print(x)
When tested on data produced by this Perl script:
$line = pack("nNccH*", 1, 2, 10, 4, '1fba');
print "$line";
The Python script yields
[1, 2, 10, 4, '1fba']
The equivalent Python function you're looking for is struct.unpack. Documentation of the format string is here: http://docs.python.org/library/struct.html
You will have a better chance of getting help if you actually explain what kind of unpacking you need. Not everyone knows Perl.