[英]Scatter plot with string X and Y coordinates
I did see this , but I was wondering if there is a way to do it without pandas. 我确实看到了 ,但是我想知道是否有没有熊猫的方法。
This is what I have tried. 这就是我尝试过的。
X_AXIS
, and Y_AXIS
are strings that are contained in xTickMarks
and yTickMarks
, respectively. X_AXIS
和Y_AXIS
是包含在字符串xTickMarks
和yTickMarks
分别。
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
. 中的每个值
X_AXIS
和Y_AXIS
具有对应Z_AXIS
value..and散射点的大小使用列表表示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()
When I run this, plt.scatter
expects a floating point value and not a string. 当我运行此命令时,
plt.scatter
需要浮点值而不是字符串。
I want to do it without pandas
. 我想没有
pandas
。
Sample values of X_AXIS, Y_AXIS and Z_AXIS: X_AXIS,Y_AXIS和Z_AXIS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]
You need numbers for the scatter plot. 您需要散点图的数字。 You can map your string values to numbers and then set the ticks to match the mapping.
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()
What I changed here compared to your code are the values for x and y: 与您的代码相比,我在这里所做的更改是x和y的值:
x = [xTickMarks.index(i) for i in X_AXIS]
y = [yTickMarks.index(i) for i in Y_AXIS]
and the ticks of your axes: 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)
Hope this helps. 希望这可以帮助。
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