问题

I am plotting a graph with a legend using the following command:

K = [1e3,1e4]
plt.plot(x[:len(y1)], y1, 'o', label=f'Simulation ($K_L$={K[i]:0e})')

which produces the following graph:

But I want the legend to be in a specific format as presented below:

K_L = 10³

K = [1e3,1e4]
plt.plot(x[:len(y1)], y1, 'o', label=f'Simulation ($K_L$={K[i]:0e})')

答案1

得分: 1

这部分内容的翻译如下：

import matplotlib.pyplot as plt

def sci_notation(number: float, n_decimals: int = 2) -> str:
    sci_notation = f"{number:e}"
    decimal_part, exponent = sci_notation.split("e")
    decimal_part = float(decimal_part)
    exponent = int(exponent)

    if decimal_part == 1.0:
        return f"$10^{exponent}$"
    else:
        return f"{decimal_part:.{n_decimals}f}x$10^{exponent}$"

K = [1e3, 1.323423e4]
plt.plot([], [], "o", label=f"Simulation ($K_L$={sci_notation(K[0])})")
plt.plot([], [], "o", label=f"Simulation ($K_L$={sci_notation(K[1])})")
plt.legend()
plt.show()

import matplotlib.pyplot as plt


def sci_notation(number: float, n_decimals: int = 2) -> str:
    sci_notation = f"{number:e}"
    decimal_part, exponent = sci_notation.split("e")
    decimal_part = float(decimal_part)
    exponent = int(exponent)

    if decimal_part == 1.0:
        return f"$10^{exponent}$"
    else:
        return f"{decimal_part:.{n_decimals}f}x$10^{exponent}$"


K = [1e3, 1.323423e4]
plt.plot([], [], "o", label=f"Simulation ($K_L$={sci_notation(K[0])})")
plt.plot([], [], "o", label=f"Simulation ($K_L$={sci_notation(K[1])})")
plt.legend()
plt.show()

答案2

得分: 1

如果您的图例已经硬编码，您可以这样做：

```python
import numpy as np
import matplotlib.pyplot as plt

K = [1e3, 1e4]  # 硬编码的图例
x = np.random.rand(100)
y1 = np.random.rand(100)

plt.plot(
  x[:len(y1)],
  y1,
  'o',
  label=f'模拟 ($K_L$=$10^{int(np.log10(K[0]))}$)'
)
plt.legend()

这里将您的 K 的 log10 值手动放在上标（指数）中。

此外，还有一个 scinot 包，可以帮助您更灵活地格式化：

import scinot as sn
a = 1.7e-2
print(sn.format(a))

1.7 × 10⁻²

<details>
<summary>英文:</summary>

If your legend is hardcoded anyway, you can do this:

import numpy as np
import matplotlib.pyplot as plt

K = [1e3,1e4] # hardcoded legend
x = np.random.rand(100)
y1 = np.random.rand(100)

plt.plot(
x[:len(y1)],
y1,
'o',
label=f'Simulation ($K_L$=$10^{int(np.log10(K[0]))}$)'
)
plt.legend()

Here the log10 of your K is put manually in the exponent (superscript).

Also, there is a `scinot` package that might help you formatting more flexible

import scinot as sn
a=1.7e-2
print(scinot.format(a))

1.7 × 10⁻²

</details>