具有字符串X和Y坐标的散点图 [英] Scatter plot with string X and Y coordinates

问题描述

我确实看到了这，但是我想知道是否有是没有熊猫的一种方式.

I did see this, but I was wondering if there is a way to do it without pandas.

这是我尝试过的. X_AXIS和Y_AXIS是分别包含在xTickMarks和yTickMarks中的字符串.

This is what I have tried. X_AXIS, and Y_AXIS are strings that are contained in xTickMarks and yTickMarks, respectively.

X_AXIS和Y_AXIS中的每个值都有一个对应的Z_AXIS值..散点的大小使用列表FREQ表示.

Each value in X_AXIS and Y_AXIS has a corresponding Z_AXIS value..and the the size of the scatter point is represented using the list 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()

运行此命令时，plt.scatter需要浮点值而不是字符串.

When I run this, plt.scatter expects a floating point value and not a string.

我想在没有pandas的情况下这样做.

I want to do it without pandas.

X_AXIS，Y_AXIS和Z_AXIS的样本值:

Sample values of X_AXIS, Y_AXIS and Z_AXIS:

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

推荐答案

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.

You need numbers for the scatter plot. You can map your string values to numbers and then set the ticks to match the mapping.

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，我在这里所做的更改是x和y的值:

What I changed here compared to your code are the values for x and y:

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

和您的坐标轴的刻度线:

and the ticks of your axes:

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)

希望这会有所帮助.

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