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python - 带有字符串 X 和 Y 坐标的散点图

转载 作者:太空宇宙 更新时间:2023-11-03 16:55:45 25 4
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我确实看到了this ,但我想知道是否有办法在没有 pandas 的情况下做到这一点。

这是我尝试过的。X_AXISY_AXIS 是分别包含在 xTickMarksyTickMarks 中的字符串。

X_AXISY_AXIS中的每个值都有一个对应的Z_AXIS值..并且使用列表来表示散点的大小频率

xTickMarks = ["0B", "1B", "2B", "1B3S", "2B2S"]
yTickMarks = ["-1S","0S", "1S", "2S", "3S", "4S"]
matplotlib.rc('font', serif='Helvetica Neue')
matplotlib.rc('text', usetex='false')
matplotlib.rcParams.update({'font.size': 10})
fig = plt.figure(figsize=(11.69,4.88)) # for landscape
axes1 = fig.add_subplot(111)

'''
Tuple of big,small cores with a tuple of power.
let x be big core
let y be small core
let z be Power
'''

plt.grid(True,linestyle='-',color='0.75')
x = X_AXIS
y = Y_AXIS
z = Z_AXIS
s = [int(FREQ[i])/1000000.0 for i in range(len(FREQ))]

plt.scatter(x,y,s=s,c=z, marker = 'o', cmap = cm.jet )
cb = plt.colorbar()
cb.set_label('Frequency', fontsize=20)

xtickNames = axes1.set_xticklabels(xTickMarks)
ytickNames = axes1.set_yticklabels(yTickMarks)
axes1.set_ylabel('Small cores')
axes1.set_xlabel('Big cores')
axes1.legend(prop={'size':5}, ncol=4)
axes1.xaxis.grid(True)
figsize=(11.69,8.27) # for landscape
fig.savefig(savepath + 'state-transition.png', bbox_inches='tight', dpi=300, pad_inches=0.1)
plt.clf()

当我运行此命令时,plt.scatter 需要浮点值而不是字符串。

我想在没有pandas的情况下做到这一点。

X_AXIS、Y_AXIS 和 Z_AXIS 的示例值:

X_AXIS = ['1B', '2B', '2B', '2B']
Y_AXIS = ['0S', '0S', '2S', '2S']
Z_AXIS = [1.5637257394958113, 1.5399805470086181, 1.4030363999998978, 1.4198133749999822]

最佳答案

您需要散点图的数字。您可以将字符串值映射到数字,然后设置刻度以匹配映射。

import matplotlib
import matplotlib.pyplot as plt
import matplotlib.cm as cm

X_AXIS = ['1B', '2B', '2B', '1B3S']
Y_AXIS = ['0S', '0S', '2S', '2S']
Z_AXIS = [1.5637257394958113, 1.5399805470086181, 1.4030363999998978, 1.4198133749999822]
FREQ = [5000000.] * 4

xTickMarks = ["0B", "1B", "2B", "1B3S", "2B2S"]
yTickMarks = ["-1S","0S", "1S", "2S", "3S", "4S"]

matplotlib.rc('font', serif='Helvetica Neue')
matplotlib.rc('text', usetex='false')
matplotlib.rcParams.update({'font.size': 10})
fig = plt.figure(figsize=(11.69,4.88)) # for landscape
axes1 = fig.add_subplot(111)

'''
Tuple of big,small cores with a tuple of power.
let x be big core
let y be small core
let z be Power
'''

plt.grid(True,linestyle='-',color='0.75')
x = [xTickMarks.index(i) for i in X_AXIS]
y = [yTickMarks.index(i) for i in Y_AXIS]
z = Z_AXIS
s = [int(FREQ[i])/1000000.0 for i in range(len(FREQ))]

plt.scatter(x,y,s=s,c=z, marker = 'o', cmap = cm.jet )
cb = plt.colorbar()
cb.set_label('Frequency', fontsize=20)

axes1.set_xlim((0, len(xTickMarks)-1))
axes1.set_ylim((0, len(yTickMarks)-1))
axes1.set_xticks(xrange(len(xTickMarks)))
axes1.set_yticks(xrange(len(yTickMarks)))
axes1.set_xticklabels(xTickMarks)
axes1.set_yticklabels(yTickMarks)
axes1.set_ylabel('Small cores')
axes1.set_xlabel('Big cores')
axes1.legend(prop={'size':5}, ncol=4)
axes1.xaxis.grid(True)
figsize=(11.69,8.27) # for landscape
#fig.savefig('state-transition.png', bbox_inches='tight', dpi=300, pad_inches=0.1)
plt.show()

与您的代码相比,我在这里更改的是 x 和 y 的值:

x = [xTickMarks.index(i) for i in X_AXIS]
y = [yTickMarks.index(i) for i in Y_AXIS]

以及轴的刻度:

axes1.set_xlim((0, len(xTickMarks)-1))
axes1.set_ylim((0, len(yTickMarks)-1))
axes1.set_xticks(xrange(len(xTickMarks)))
axes1.set_yticks(xrange(len(yTickMarks)))
axes1.set_xticklabels(xTickMarks)
axes1.set_yticklabels(yTickMarks)

enter image description here

希望这有帮助。

关于python - 带有字符串 X 和 Y 坐标的散点图,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35432918/

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