Python 具有字符串X和Y坐标的散点图

我确实看到了，但我想知道是否有一种方法可以在没有熊猫的情况下做到这一点

这就是我尝试过的。

X_轴

和

Y_轴

分别是

xTickMarks

和

yTickMarks

中包含的字符串

X_轴

和

Y_轴

中的每个值都有一个相应的

Z_轴

值。散射点的大小使用列表

FREQ

表示

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

需要一个浮点值而不是字符串

我想在没有熊猫的情况下做这件事

X_轴、Y_轴和Z_轴的样本值：

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

---

散点图需要数字。您可以将字符串值映射为数字，然后设置记号以匹配映射

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)

---

希望这能有所帮助。

请添加X、Y和Z轴值的示例欢迎：-）您的刻度标签正确，但刻度数错误。