请考虑以下代码:
from numpy import log2
import matplotlib.pyplot as plt
xdata = [log2(x)*(10/log2(10)) for x in range(1,11)]
ydata = range(10)
plt.plot(xdata, ydata)
plt.show()
这会生成以下图:我的问题是,如何修改它,以便绘图与输入的数据完全相同,显示为直线?这基本上需要适当地缩放x轴,但我无法想象如何做到这一点。这样做的原因是我显示的函数在开始时变化很小,但在有效间隔结束时开始更多地波动,所以我希望在结束时有更高的水平分辨率。如果有人可以为我的方法提出替代解决方案,请随意这样做!
答案 0 :(得分:7)
这是how完成的。一个好的example可以跟随。你只是继承ScaleBase类。
这是你的转变。当你削减所有自定义格式化程序和东西时,它并不太复杂。只是有点冗长。
from numpy import log2
import matplotlib.pyplot as plt
from matplotlib import scale as mscale
from matplotlib import transforms as mtransforms
class CustomScale(mscale.ScaleBase):
name = 'custom'
def __init__(self, axis, **kwargs):
mscale.ScaleBase.__init__(self)
self.thresh = None #thresh
def get_transform(self):
return self.CustomTransform(self.thresh)
def set_default_locators_and_formatters(self, axis):
pass
class CustomTransform(mtransforms.Transform):
input_dims = 1
output_dims = 1
is_separable = True
def __init__(self, thresh):
mtransforms.Transform.__init__(self)
self.thresh = thresh
def transform_non_affine(self, a):
return 10**(a/10)
def inverted(self):
return CustomScale.InvertedCustomTransform(self.thresh)
class InvertedCustomTransform(mtransforms.Transform):
input_dims = 1
output_dims = 1
is_separable = True
def __init__(self, thresh):
mtransforms.Transform.__init__(self)
self.thresh = thresh
def transform_non_affine(self, a):
return log2(a)*(10/log2(10))
def inverted(self):
return CustomScale.CustomTransform(self.thresh)
mscale.register_scale(CustomScale)
xdata = [log2(x)*(10/log2(10)) for x in range(1,11)]
ydata = range(10)
plt.plot(xdata, ydata)
plt.gca().set_xscale('custom')
plt.show()
答案 1 :(得分:2)
最简单的方法是使用semilogy
from numpy import log2
import matplotlib.pyplot as plt
xdata = log2(range(1,11)) * (10/log2(10))
ydata = range(10)
plt.semilogy(xdata, ydata)
plt.show()