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我希望注释中的数字像 {x}\cdot10^{y}，而不是现在的：{x}E+{y}。是否有合适的字符串格式化程序？

对我来说最理想的解决方案是使用格式字符串，如“%.2e”，但会自动转换为十次方表示。

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3 回答 3

14

您可以定义自己的字符串格式化程序以用于 LaTeX 或 Mathtext。在sci_notation()下面定义的函数中，您可以指定有效小数位数以及要打印的小数位数（默认为有效小数位数）。也可以明确指定应该使用哪个指数。

from math import floor, log10

# Use LaTeX as text renderer to get text in true LaTeX
# If the two following lines are left out, Mathtext will be used
import matplotlib as mpl
mpl.rc('text', usetex=True)

import matplotlib.pyplot as plt

# Define function for string formatting of scientific notation
def sci_notation(num, decimal_digits=1, precision=None, exponent=None):
    """
    Returns a string representation of the scientific
    notation of the given number formatted for use with
    LaTeX or Mathtext, with specified number of significant
    decimal digits and precision (number of decimal digits
    to show). The exponent to be used can also be specified
    explicitly.
    """
    if exponent is None:
        exponent = int(floor(log10(abs(num))))
    coeff = round(num / float(10**exponent), decimal_digits)
    if precision is None:
        precision = decimal_digits

    return r"${0:.{2}f}\cdot10^{{{1:d}}}$".format(coeff, exponent, precision)

scinum = -3.456342e-12

# Annotation with exponent notation using `e` as separator
plt.annotate(scinum, (0.5,0.5), ha='center', fontsize=20)

# Annotation with scientific notation using `\cdot 10` as separator
plt.annotate(sci_notation(scinum,1), (0.5,0.4), ha='center', fontsize=20)

# Annotation with scientific notation using `\cdot 10` as separator
# with 1 significant decimal digit and 2 decimal digits shown as well
# as a given exponent.
plt.annotate(sci_notation(scinum,1,2,exponent=-14), (0.5,0.3), ha='center', fontsize=20)

plt.title('Scientific notation', fontsize=14)

plt.show()

在此处输入图像描述

于 2013-08-19T12:25:06.970 回答
1 
def round_to_n(x, n):
    " Round x to n significant figures "
    return round(x, -int(py.floor(py.sign(x) * py.log10(abs(x)))) + n)

def str_fmt(x, n=2):
    " Format x into nice Latex rounding to n"
    power = int(py.log10(Round_To_n(x, 0)))
    f_SF = Round_To_n(x, n) * pow(10, -power)
    return r"${}\cdot 10^{}$".format(f_SF, power)

>>> x = 1203801.30201
>>> str_fmt(x)
$1.2\\cdot 10^6$

如何参数化存在许多变化，例如，您可以指定指数 ( y) 而不是自动生成它，但原理保持不变。

于 2013-08-19T11:42:45.447 回答
0

这是一种相当简单的方法，尽管它可能需要根据指数中的位数进行调整：

x=17875.764
print(x)
print(((str("%.1e" % x)).replace("e", ' \\cdot 10^{ ')).replace("+0", "") + ' } ')
于 2021-10-26T15:21:37.487 回答