格式化具有变量和LaTeX的图表标题

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英文:

Formating the tittle of a graph that has varabiles and latex

问题

使用以下代码中的用户输入来更改图表标题,我想让它以最简单的形式显示图表标题,例如当a=1,b=0,c=-1时,我希望它显示为 "x-1",但实际上显示的是 "1x+0x+-1"。

英文:

i was trying to make a graph tittle change depending on the users input using the following piece of code plt.title(f"{a}$x^2+{b}x+{c}$")

i would like it to display the title of the graph in the most simple form for example when a=1 b=0 c=-1

i want it to display x-1 as the title

but instead it is displaying 1x+0x+-1 as the tittle

答案1

得分: 1

a = 1
b = 0
c = -1

def format_title(a, b, c):
    title = ''
    if a == 1:
        title += 'x^2'
    else:
        title += f'{a}x^2'
    if b != 0:
        if b > 0:
            title += '+'
        title += f'{b}x'
    if c != 0:
        if c > 0:
            title += '+'
        title += f'{c}'
    return title

print(format_title(a, b, c))

Output:

x^2-1
英文:

Simple python code:

a = 1
b = 0
c = -1


def format_title(a, b, c):
    title = ''
    if a == 1:
        title += 'x^2'
    else:
        title += f'{a}x^2'
    if b != 0:
        if b > 0:
            title += '+'
        title += f'{b}x'
    if c != 0:
        if c > 0:
            title += '+'
        title += f'{c}'
    return title


 print(format_title(a, b, c))

Output:

x^2-1

答案2

得分: 0

以下是代码的翻译部分:

def ft(*coefs, exponent="x", cdp=0):
    title = ""
    cstring = f"{{:.{cdp}f}}"  # 用于格式化系数的小数位数
    nc = len(coefs) - 1
    for i in range(nc, -1, -1):
        if coefs[nc - i] != 0:
            # 获取符号
            sign = "+" if coefs[nc - i] > 0 and i < nc else ("-" if i == nc and coefs[nc - i] == -1 else "")
            
            # 获取系数
            c = cstring.format(coefs[nc - i]) if abs(coefs[nc - i]) != 1 or i < nc else ""
            
            # 获取指数
            exp = f"{exponent}^{i}" if i > 1 else (exponent if i == 1 else "")
            part = f"{sign}{c}{exp}"
            title += f"{part}"
    
    return f"${title}$"

这段代码是一个通用的函数,它可以接受任意数量的系数(不仅限于二次方程),并可以将系数输出为指定小数位数(默认为不带小数位,即整数)。对于你的例子,你可以使用:

print(ft(1, 0, -1))
$x^2-1$

但你也可以使用以下方式:

print(ft(-4.765, 0, 2.383, -0.351, cdp=2))
$-4.76x^3+2.38x-0.35$
英文:

A more complicated, but generalisable option would be:

def ft(*coefs, exponent=&quot;x&quot;, cdp=0):
    title = &quot;&quot;
    cstring = f&quot;{{:.{cdp}f}}&quot;  # number of decimal places to format the coefficents
    nc = len(coefs) - 1
    for i in range(nc, -1, -1):
        if coefs[nc - i] != 0:
            # get the sign
            sign = &quot;+&quot; if coefs[nc - i] &gt; 0 and i &lt; nc else (&quot;-&quot; if i == nc and coefs[nc - i] == -1 else &quot;&quot;)
            
            # get the coefficient
            c = cstring.format(coefs[nc - i]) if abs(coefs[nc - i]) != 1 or i &lt; nc else &quot;&quot;
            
            # get the exponent
            exp = f&quot;{exponent}^{i}&quot; if i &gt; 1 else (exponent if i == 1 else &quot;&quot;)
            part = f&quot;{sign}{c}{exp}&quot;
            title += f&quot;{part}&quot;
    
    return f&quot;${title}$&quot;

This can take in any number of coefficients (so not just quadratics) and also output the coefficients with to any number of decimals places (it defaults to just output with no decimal places, i.e., as integers). For you case you still have:

print(ft(1, 0, -1))
$x^2-1$

but you could also do, e.g.,

print(ft(-4.765, 0, 2.383, -0.351, cdp=2))
$-4.76x^3+2.38x-0.35$

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  • 本文由 发表于 2023年3月4日 00:26:32
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