CODEWITHSUNDEEP
pythonformattinglatex
pafcu
pafcu

Reputation: 8128

Format number using LaTeX notation in Python

Using format strings in Python I can easily print a number in "scientific notation", e.g.

>> print '%g'%1e9
1e+09

What is the simplest way to format the number in LaTeX format, i.e. 1\times10^{+09}?

Upvotes: 17

Views: 12305

Answers (4)

Giuseppe499
Giuseppe499

Reputation: 11

I wrote a super simple python3 package called real2tex.

You can install it with pip:

pip install real2tex

And use it as follows:

from real2tex import real2tex

tex = real2tex(1.2345e-6, precision=2)
print(tex) # "1.23 \cdot 10^{\minus 6}"

If you are interested in the implementation, here I explain how it works. Firtst we have a function that takes a real number and returns a tuple representing it in scientific notation. It returns the mantissa, the exponent, and a boolean indicating if the exponent is negative. To do this, I exploit the built-in format specifier "{:.{precision}e}" that converts the number to a string representing it in scientific notation. The variable precision controls the number of decimal places in the mantissa (e.g., if precision=2, it will return something like "1.23e-6"). Once i have this string, I parse it to extract the mantissa, the exponent, and the sign of the exponent:

def scientific_notation(num: Real, precision: int = 2) -> tuple[str, str, bool]:
    num_str = f"{num:.{precision}e}"
    mantissa, exponent = num_str.split("e")
    negative_exponent = True if exponent[0] == "-" else False
    exponent = exponent[1:]
    # Remove leading zeros in the exponent
    exponent = exponent.lstrip("0")
    if len(exponent) == 0:
        exponent = "0"
    # Remove trailing zeros in the mantissa
    mantissa = mantissa.rstrip("0")
    # Remove trailing dot in the mantissa
    mantissa = mantissa.rstrip(".")
    # Remove sign if mantissa is zero
    if mantissa == "-0":
        mantissa = "0"
    return mantissa, exponent, negative_exponent

Once I have the mantissa, the exponent, and the sign of the exponent, it is easy to convert this information to LaTeX format:

def real2tex(
    num: Real, precision: int = 2, multiply_symbol: str = "\\cdot", no_10_to_the_zero: bool = True
) -> str:
    mantissa, exponent, negative_exponent = scientific_notation(num, precision)
    if negative_exponent:
        exponent = f"\\minus {exponent}"
    if exponent == "0" and no_10_to_the_zero:
        return f"{mantissa}"
    return f"{mantissa} {multiply_symbol} 10^{{{exponent}}}"

If you want to edit the code or report an issue, you can find it on GitHub.

Upvotes: 1

MountainDrew
MountainDrew

Reputation: 473

Install num2tex:

pip install num2tex

and use it as so:

>>> from num2tex import num2tex
>>> '{:.0e}'.format(num2tex(1e9))
'1 \\times 10^{9}'

num2tex inherits from str so the format function can be used in the same way.

You can also change the format of the exponent by using num2tex.configure() (adding this in response to @Matt's comment).

>>>from num2tex import num2tex
>>>from num2tex import configure as num2tex_configure
>>>num2tex_configure(exp_format='cdot')
>>>num2tex(1.3489e17)
'1.3489 \cdot 10^{17}'
>>>num2tex_configure(exp_format='parentheses')
'1.3489 (10^{17})'

As of now this is undocumented in the GitHub, I'll try to change this soon!

Disclaimer: After using (and upvoting) Lauritz V. Thaulow's answer for a while (for Jupyter, Matplotlib etc.) I thought it would be better for my workflow to write a simple Python module, so I created num2tex on GitHub and registered it on PyPI. I would love to get some feedback on how to make it more useful.

Upvotes: 8

Lauritz V. Thaulow
Lauritz V. Thaulow

Reputation: 50985

The siunitx LaTeX package solves this for you by allowing you to use the python float value directly without resorting to parsing the resulting string and turning it into valid LaTeX.

>>> print "\\num{{{0:.2g}}}".format(1e9)
\num{1e+09}

When the LaTeX document is compiled, the above code will be turned into enter image description here. As andybuckley points out in the comments, the plus sign might not be accepted by siunitx (I've not tested it), so it may be necessary to do a .repace("+", "") on the result.

If using siunitx is somehow off the table, write a custom function like this:

def latex_float(f):
    float_str = "{0:.2g}".format(f)
    if "e" in float_str:
        base, exponent = float_str.split("e")
        return r"{0} \times 10^{{{1}}}".format(base, int(exponent))
    else:
        return float_str

Testing:

>>> latex_float(1e9)
'1 \\times 10^{9}'

Upvotes: 29

ecatmur
ecatmur

Reputation: 157314

You can write a frexp10 function:

def frexp10(x):
    exp = int(math.floor(math.log10(abs(x))))
    return x / 10**exp, exp

Formatting in LaTeX style is then:

'{0}^{{{1:+03}}}'.format(*frexp10(-1.234e9))

Upvotes: 6

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