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带字符串X和Y坐标的散点图

scatter matplotlib numpy python

tandem · 技术社区 · 9 年前

我确实看到了 this ,但我想知道有没有办法在没有熊猫的情况下做到这一点。

这是我尝试过的。 X_AXIS 和 Y_AXIS 是包含在 xTickMarks 和 yTickMarks 分别地

中的每个值 X轴(_A) 和 Y轴 具有相应的 Z_AXIS 值..,分散点的大小用列表表示 FREQ .

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()

当我运行这个时, plt.scatter 应为浮点值,而不是字符串。

我想不用 pandas .

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]

1 回复 | 直到 8 年前

Julien Spronck 9 年前

散点图需要数字。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]

和你的轴的滴答声:

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)

希望这有帮助。