Python Numpy:将散点图转换为二维阵列
假设我有以下几点:Python Numpy:将散点图转换为二维阵列,python,numpy,matplotlib,scipy,Python,Numpy,Matplotlib,Scipy,假设我有以下几点: import pyplot as plt import numpy as np '''array([[29, 13, 11, 4, 5], #dataMag [19, 16, 25, 9, 10], [16, 22, 14, 18, 26], [ 9, 17, 8, 9, 777]]) array([[205, 338, 380, 428, 228], #dataX [199, 546, 430, 95,
import pyplot as plt
import numpy as np
'''array([[29, 13, 11, 4, 5], #dataMag
[19, 16, 25, 9, 10],
[16, 22, 14, 18, 26],
[ 9, 17, 8, 9, 777]])
array([[205, 338, 380, 428, 228], #dataX
[199, 546, 430, 95, 374],
[418, 85, 260, 236, 241],
[308, 481, 133, 136, 83]])
array([[ 0.48, 0.83, 0.71, 0.12, 0.], #dataY
[ 0.09, 0. , 0.7 , 0.43, 0.54],
[ 0.58, 0. , 0.56, 0.18, 0.25],
[ 0.96, 0.26, 0.57, 0. , 0.82])'''
plt.scatter(x=dataX.flat, y=dataY.flat, c=dataMag.flat, vmin=np.min(dataMag),
vmax=np.max(dataMag), marker='s', cmap='hot')
plt.show()
这给了我以下结果:
在Numpy(或Scipy等)中是否有方法将它们表示为(a,b)2D数组,而不是使用三个数组来获得2D图像?编辑我在下面保留了我的原始答案,但深入研究了您之前关于同一主题的问题,代码如下所示。请注意,它不会处理重复的值,因此,如果为同一位置分配了多个值,则只会保留其中一个值。而且,这会弄乱散点图的比例,所以像我的原始答案这样的东西可能更适合你所追求的。但无论如何,代码如下:
x_, x_idx = np.unique(np.ravel(dataX), return_inverse=True)
y_, y_idx = np.unique(np.ravel(dataY), return_inverse=True)
newArray = np.zeros((len(x_), len(y_)), dtype=dataMag.dtype)
newArray[x_idx, y_idx] = np.ravel(dataMag)
>>> newArray
array([[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 777, 0, 0],
[ 22, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0],
[ 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 19, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 29, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 18, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 26, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 14, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16, 0, 0, 0, 0, 0],
[ 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 25, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 17, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
原始答案
如果dataX
和dataY
都是整数数组,那么实现将非常简单。但因为它们似乎不一定是,所以您需要进行一些舍入,为此,您需要首先为阵列在每个方向上选择步长,然后您可以执行以下操作:
from __future__ import division
x_step, y_step = 25, 0.10
x = np.round(dataX / x_step).astype(int)
y = np.round(dataY / y_step).astype(int)
x_m, x_M = np.min(x), np.max(x)
y_m, y_M = np.min(y), np.max(y)
newArray = np.zeros((x_M - x_m + 1, y_M - y_m + 1), dtype=dataMag.dtype)
newArray[x - x_m, y - y_m] = dataMag
>>> newArray
array([[ 22, 0, 0, 0, 0, 0, 0, 0, 777, 0, 0],
[ 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0],
[ 9, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 19, 0, 0, 0, 29, 0, 0, 0, 0, 0],
[ 5, 0, 18, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 26, 0, 0, 0, 14, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 13, 0, 0],
[ 0, 0, 0, 0, 0, 10, 0, 11, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 4, 0, 0, 0, 0, 16, 25, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 17, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[ 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
在执行此操作时,必须小心,确保舍入步长足够小,以便不会有两个值舍入到数组中的同一位置,否则会丢失信息。例如:
x_step, y_step = 50, 0.10
...
>>> newArray
array([[ 22, 0, 0, 0, 9, 0, 0, 0, 777, 0, 0],
[ 9, 0, 0, 0, 0, 0, 8, 0, 0, 0, 0],
[ 0, 19, 0, 0, 0, 29, 0, 0, 0, 0, 0],
[ 5, 0, 26, 0, 0, 0, 14, 0, 0, 0, 0],
[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9],
[ 0, 0, 0, 0, 0, 10, 0, 0, 13, 0, 0],
[ 0, 0, 0, 0, 0, 0, 16, 11, 0, 0, 0],
[ 0, 4, 0, 0, 0, 0, 0, 25, 0, 0, 0],
[ 0, 0, 0, 17, 0, 0, 0, 0, 0, 0, 0],
[ 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
在位置
[3,2]
中,只显示26,而不是前面示例中相应单元格中的18和26。您希望该数组的结构是什么?你是说你想要一个有点的1和没有点的0的2D数组,还是什么?我不知道你想如何在一个2D数组中变换这三个数组,但是你可以很容易地得到一个3d数组。考虑您的数组名为“代码> MAG< <代码> >代码> >代码< y>代码>,然后<代码> ALLMES=dStad((x.rVEVER),y.rVIEW(),MAG.RVELL())< < /代码>和<代码>散射(x= Alel[[…,0 ],y= Alel[[…,1 ],c= Alel-[…,2 ])< /C> >,将是一个数组,例如“代码> NeWraye< /Cord>,每个轴<代码> [i,j]是来自dataX
的有序点,dataY
由来自dataMag
的相应量级组成这是一个视觉效果(也是我要求的,但没有用)。嗯……我真诚地感谢你的帮助,但我不认为这是我想要的。例如:对于dataMag[3,4]=777
,值应重新映射到newArray[0,14]
,因为dataX[3,4]=83
(dataX
中的第一个排序元素)和dataY[3,4]=0.82
(包括0的dataY
中的第14个排序元素)。你能解释一下你的想法吗。什么是\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。来自未来导入除法的是为了避免整数的除法被向下舍入,例如,没有导入3/2==1
和导入3/2=1.5
。我的天啊……我自由了!非常感谢你。我与这个愚蠢问题斗争的日子结束了,我终于可以进步了!为了满足您的好奇心,绘图只是我需要的数据功能之一。对于其他阵列,需要二维阵列。