带上标指数的科学记数法

Question

我正在尝试使用以 10 为底的指数以科学记数法格式化数字，例如使用 python 3 将 0.00123 写为 1.23x10^–3。

我发现这个很棒的函数可以打印 1.23x10^-3，但是如何用上标替换插入符号指数？

def sci_notation(number, sig_fig=2):
    ret_string = "{0:.{1:d}e}".format(number, sig_fig)
    a,b = ret_string.split("e")
    b = int(b)         # removed leading "+" and strips leading zeros too.
    return a + "x10^" + str(b)

print(sci_notation(0.001234, sig_fig=2))      # Outputs 1.23x10^-3

该函数修改自https://stackoverflow.com/a/29261252/8542513。

我尝试合并 https://stackoverflow.com/a/8651690/8542513 的答案来格式化上标，但我不确定 sympy 如何处理变量：

from sympy import pretty_print as pp, latex
from sympy.abc import a, b, n

def sci_notation(number, sig_fig=2):
  ret_string = "{0:.{1:d}e}".format(number, sig_fig)
  a,b = ret_string.split("e")
  b = int(b)             #removed leading "+" and strips leading zeros too.
  b = str(b)
  expr = a + "x10"**b    #Here's my problem
  pp(expr)               # default
  pp(expr, use_unicode=True)
  return latex(expr)

print(latex(sci_notation(0.001234, sig_fig=2))) 

这将返回：TypeError: unsupported operand type(s) for ** or pow(): 'str' and 'int'

Answer 1

这是一个简单的解决方案：

def SuperScriptinate(number):
  return number.replace('0','⁰').replace('1','¹').replace('2','²').replace('3','³').replace('4','⁴').replace('5','⁵').replace('6','⁶').replace('7','⁷').replace('8','⁸').replace('9','⁹').replace('-','⁻')

def sci_notation(number, sig_fig=2):
    ret_string = "{0:.{1:d}e}".format(number, sig_fig)
    a,b = ret_string.split("e")
    b = int(b)         # removed leading "+" and strips leading zeros too.
    return a + "x10^" + SuperScriptinate(str(b))

Answer 2

scinot 包以科学计数法格式化数字

import scinot as sn
a=4.7e-8
print(scinot.format(a))

4.7 × 10⁻⁸

或在字符串中

print('The value is {0:s}'.format(scinot.format(a)))

The value is 4.7 × 10⁻⁸

Answer 3

我认为你的主要问题是

how can the caret-exponent be replaced with a superscript?

如果您在 Jupyter 笔记本中使用 python，有一个简单的方法：

  from IPython.display import display, Math, Latex

  # if the number is in scientific format already
  display(Math('2.14e-6'.replace('e', r'\times 10^{') + '}'))

  # if it is not:
  d = "%e" % (number)
  # then use the above form: display(Math(d.replace('e', r'\times ...