I am using ipython 5.8.0 on Debian 10.
This is how output looks like:
In [1]: 50*50
Out[1]: 2500
Is it possible to configure ipython to print all numbers with thousands separators? ie:
In [1]: 50*50
Out[1]: 2'500
In [2]: 5000*5000
Out[2]: 25'000'000
And perhaps, is it possible to make ipython also understand thousands separators on input?
In [1]: 5'000*5'000
Out[1]: 25'000'000
UPDATE
The accepted answer from #Chayim Friedman works for integers, but does not work for float:
In [1]: 500.1*500
Out[1]: 250050.0
Also, when it works, it uses , as the character for thousand separator:
In [1]: 500*500
Out[1]: 250,000
Can I use ' instead?
Using ' as thousands separator in input is quite problematic because Python uses ' to delimit strings, but you can use _ (PEP 515, Underscores in Numeric Literals):
Regarding output, this is slightly harder, but can be done using IPython extensions.
Put the following Python code in a new file at ~/.ipython/extensions/thousands_separator.py:
default_int_printer = None
def print_int(number, printer, cycle):
printer.text(f'{number:,}') # You can use `'{:,}'.format(number)` if you're using a Python version older than 3.6
def load_ipython_extension(ipython):
global default_int_printer
default_int_printer = ipython.display_formatter.formatters['text/plain'].for_type(int, print_int)
def unload_ipython_extension(ipython):
ipython.display_formatter.formatters['text/plain'].for_type(int, default_int_printer)
This code tells IPython to replace the default int formatter with one that prints thousand separators when this extension is loaded, and restore the original when it is unloaded.
Edit: If you want a different separator, for instance ', replace the f'{number:,}' with f'{number:,}'.replace(',', "'").
You can load the extension using the magic command %load_ext thousands_separator and unload it using %unload_ext thousands_separator, but if you want it always, you can place it in the default profile.
Run the following code in the terminal:
ipython3 profile create
It will report that a file ~/.ipython/profile_default/ipython_config.py was created. Enter it, and search for the following string:
## A list of dotted module names of IPython extensions to load.
#c.InteractiveShellApp.extensions = []
Replace it with the following:
# A list of dotted module names of IPython extensions to load.
c.InteractiveShellApp.extensions = [
'thousands_separator'
]
This tells IPython to load this extension by default.
Done!
Edit: I saw that you want to a) use ' as separator, and b) do the same for floats:
Using different separator is quite easy: just str.replace():
def print_int(number, printer, cycle):
printer.text(f'{number:,}'.replace(',', "'"))
Doing the same for floats is also easy: just setup print_int so it prints floats to. I also suggest to change the name to print_number.
Final code:
default_int_printer = None
default_float_printer = None
def print_number(number, printer, cycle):
printer.text(f'{number:,}'.replace(',', "'"))
def load_ipython_extension(ipython):
global default_int_printer
global default_float_printer
default_int_printer = ipython.display_formatter.formatters['text/plain'].for_type(int, print_number)
default_float_printer = ipython.display_formatter.formatters['text/plain'].for_type(float, print_number)
def unload_ipython_extension(ipython):
ipython.display_formatter.formatters['text/plain'].for_type(int, default_int_printer)
ipython.display_formatter.formatters['text/plain'].for_type(float, default_float_printer)
After update: you can subclass int:
class Int(int):
def __repr__(self):
return "{:,}".format(self)
Int(1000)
# 1,000
I don't believe you can achieve all that you are looking for without rewriting the iPython interpreter, which means changing the Python language specification, to be able to input numbers with embedded ' characters and have them ignored. But you can achieve some of it. Subclassing the int class is a good start. But you should also overload the various operators you plan on using. For example:
class Integer(int):
def __str__(self):
# if you want ' as the separator:
return "{:,}".format(self).replace(",", "'")
def __add__(self, x):
return Integer(int(self) + x)
def __mul__(self, x):
return Integer(int(self) * x)
"""
define other operations: __sub__, __floordiv__, __mod__, __neg__, etc.
"""
i1 = Integer(2)
i2 = Integer(1000) + 4.5 * i1
print(i2)
print(i1 * (3 + i2))
Prints:
1'009
2'024
Update
It seems that for Python 3.7 you need to override the __str__ method rather than the __repr__ method. This works for Python 3.8 and should work for later releases as well.
Update 2
import locale
#locale.setlocale(locale.LC_ALL, '') # probably not required
print(locale.format_string("%d", 1255000, grouping=True).replace(",", "'"))
Prints:
1'255'000
An alternative if you have package Babel from the PyPi repository:
from babel import Locale
from babel.numbers import format_number
locale = Locale('en', 'US')
locale.number_symbols['group'] = "'"
print(format_number(1255000, locale='en_US'))
Prints:
1'255'000
Or if you prefer to custom-tailor a locale just for this purpose and leave the standard en_US locale unmodified. This also shows how you can parse input values:
from copy import deepcopy
from babel import Locale
from babel.numbers import format_number, parse_number
my_locale = deepcopy(Locale('en', 'US'))
my_locale.number_symbols['group'] = "'"
print(format_number(1255000, locale=my_locale))
print(parse_number("1'125'000", locale=my_locale))
Prints:
1'255'000
1125000
Based on PEP-0378, you can use the following code:
a = 1200
b = 500
c = 10
#res = a
#res = a*b
res = a*b*c
dig = len(str(res)) # to figure out how many digits are required in result
print(format(res, "{},d".format(dig)))
It will produce:
6,000,000
I have a list of regex expressions that I want to find in certain docs.
x = ['\bin\sapp\sdata\b','\bin\sapp\sdata\b','\benough\sdata\b']
The patterns repeat themselves so I converted them to a set (see the first and second values in the list)
y = set(x)
When I try to find them in a specific doc it doesn't find them since it doesn't take them as a repr version:
import pandas as pd
import re
results = list()
doc = 'they wanted in app data and we did not provide it'
for value in y:
results.append(re.findall(pattern = value,string=doc))
results = list(filter(None, results))
results
How do I overcome this?
Thanks
The problem was with the python 3.7 version. The error I got was "bad escape \l at position 0" Once I changed the re to regex it worked perfectly fine, even with the "messed up coding
Is there a way to calculate log base c in python?
c is a variable and may change due to some dependencies.
I am new to programming and also python3.
There is already a built in function in the math module in python that does this.
from math import log
def logOc(c, num):
return log(num,c)
print(log(3,3**24))
You can read more about log and the python math module here
Yes, you can simply use math's function log():
import math
c = 100
val = math.log(10000,c) #Where the first value is the number and the second the base.
print(val)
Example:
print(val)
2.0
I have an array of complex numbers in Matlab and I want to import that data in Python. I have tried all methods including Scipy module and h5py etc. Can anyone tell me any other possible way?
My Matlab version is 2017b. and python version is 2.7.
In MATLAB, save your data with the '-v7' option:
myMat = complex(rand(4), rand(4));
save('myfile', 'myMat', '-v7')
In Python, load the .mat file with scipy.io.loadmat. The result is a Python dict:
>>> d = scipy.io.loadmat('myfile.mat')
>>> m = d['myMat']
>>> m[0,0]
'(0.421761282626275+0.27692298496088996j)'
and so on.
def shufflemode():
import random
combined = zip(question, answer)
random.shuffle(combined)
question[:], answer[:] = zip(*combined)
but then i get the error:
TypeError: object of type 'zip' has no len()
What do I do im so confused
I wonder the same thing. According to:
randomizing two lists and maintaining order in python
You should be able to do it like the OP tried, but i also get the same error. I think the ones from the link are using python 2 and not 3, could this be the problem?
This is an issue between Python 2 and Python 3. In Python 2 using shuffle after zip works, because zip returns a list. In Python 3: "TypeError: object of type 'zip' has no len()" because zip returns an iterator in Python 3.
Solution, use list() to convert to a list:
combined = list(zip(question, answer))
random.shuffle(combined)
The error appeared with shuffle() because shuffle() uses len().
References issue:
The zip() function in python 3
Stumbled upon this and was surprised to learn about the random.shuffle method. So I tried your example, and it worked for me in Python 2.7.5:
def shufflemode():
import random
combined = zip(question, answer)
random.shuffle(combined)
question[:], answer[:] = zip(*combined)
question = ["q1","q2","q3","q4","q5"]
answer = ["a1","a2","a3","a4","a5"]
if __name__ == "__main__":
shufflemode()
print question,answer
The result is two lists with the same randomized sequence of questions and answers*strong text*:
>>>
['q3', 'q2', 'q5', 'q4', 'q1'] ['a3', 'a2', 'a5', 'a4', 'a1']
>>>