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Here is what I am trying to do. I have the following code:

# -*- coding: utf-8 -*- from pylab import * import matplotlib.pyplot as plt import numpy as np ### processing function def store(var,textFile): data=loadtxt(textFile,skiprows=1) it=[] eps=[] sig=[] tc=[] sc=[] te=[] se=[] ubf=[] for i in range(0,len(data)): it.append(float(data[i,1])) eps.append(float(data[i,0])) sig.append(float(data[i,4])) tc.append(float(data[i,6])) sc.append(float(data[i,2])) te.append(float(data[i,7])) se.append(float(data[i,3])) ubf.append(float(data[i,8])) var.append(it) var.append(eps) var.append(sig) var.append(tc) var.append(sc) var.append(te) var.append(se) var.append(ubf) ### data input dataFile1='0101005_5k_tensionTestCentreCrack_l0.001a0_r0.02' a1=[] store(a1,dataFile1) dataFile2='0101005_10k_tensionTestCentreCrack_l0.001a0_r0.02' a2=[] store(a2,dataFile2) dataFile3='0101005_20k_tensionTestCentreCrack_l0.001a0_r0.01' a3=[] store(a3,dataFile3) dataFile4='0101005_40k_tensionTestCentreCrack_l0.001a0_r0.02' a4=[] store(a4,dataFile4) dataFile5='0101005_80k_tensionTestCentreCrack_l0.001a0_r0.02' a5=[] store(a5,dataFile5) dataFile6='0101005_120k_tensionTestCentreCrack_l0.001a0_r0.02' a6=[] store(a6,dataFile6) dataFile7='0101005_200k_tensionTestCentreCrack_l0.001a0_r0.02' a7=[] store(a7,dataFile7) ### plot control rcParams.update({'legend.numpoints':1,'font.size': 20,'axes.labelsize':25,'xtick.major.pad':10,'ytick.major.pad':10,'legend.fontsize':20}) lw=2 ms=10 ### plots #savefig(dataFile1+'_sigVSeps.eps',dpi=1000,format='eps',transparent=False) hcl=0.005 ###plot of fracture toughness vs. square root of mean radius A=[0.0023**0.5, 0.0019**0.5, 0.0015**0.5, 0.0012**0.5, 0.0009**0.5, 0.0008**0.5, 0.0007**0.5] B=[max(a1[2])*((pi*hcl)**0.5) ,max(a2[2])*((pi*hcl)**0.5) ,max(a3[2])*((pi*hcl)**0.5), max(a4[2])*((pi*hcl)**0.5), max(a5[2])*((pi*hcl)**0.5), max(a6[2])*((pi*hcl)**0.5), max(a7[2])*((pi*hcl)**0.5)] #B=[max(a1[2]),max(a2[2]),max(a3[2]),max(a4[2])] figure(4,figsize=(12,10)) grid() xlabel('$\sqrt{R} [m]$') ##axis(xmin=0,xmax=0.1) plot(A,[x/1e6 for x in B],'-ko',linewidth=lw) for xy in zip(A,[x/1e6 for x in B]): annotate('(%s, %s)' % xy, xy=xy, textcoords='data') ylabel(r'$K_{Ic} [MPa.\sqrt{m}]$') title(r'Fracture toughness $(K_{Ic})$ as a function of square root of mean particle radius $\sqrt{R}$', fontsize=14, color='blue') G=[(1.42*1e-5, 8.5*1e-2), (1.19*1e-5, 7.8*1e-2), (1.03*1e-5, 6*1e-2), (8.95*1e-6, 4.7*1e-2), (7.63*1e-6, 3.8*1e-2), (7.12*1e-6, 3.2*1e-2), (5.72*1e-6, 2.6*1e-2)] PN=[5*1e3, 10*1e3, 20*1e3, 40*1e3, 80*1e3, 120*1e3, 200*1e3] figure(5,figsize=(12,10)) for PNe, Ge, in zip(PN, G): scatter([PNe]*len(Ge), Ge, color=['blue', 'green']) grid() xlim(xmin=0, xmax=200000) #ylim(ymin=0, ymax=1) xlabel('Number of particles') ylabel(r'Energy release rate') figure(6,figsize=(12,10)) ax1=subplot(1,1,1) grid() xlabel(r'$\varepsilon_1$ [millistrain]') #axis(xmin=0,xmax=0.12) ax1.plot([x*1e3 for x in a1[1]],[x/1e6 for x in a1[2]],'-b',linewidth=lw) ax1.plot([x*1e3 for x in a2[1]],[x/1e6 for x in a2[2]],'-g',linewidth=lw) ax1.plot([x*1e3 for x in a3[1]],[x/1e6 for x in a3[2]],'-r',linewidth=lw) ax1.plot([x*1e3 for x in a4[1]],[x/1e6 for x in a4[2]],'-y',linewidth=lw) ax1.plot([x*1e3 for x in a5[1]],[x/1e6 for x in a5[2]],'-m',linewidth=lw) ax1.plot([x*1e3 for x in a6[1]],[x/1e6 for x in a6[2]],'-c',linewidth=lw) ax1.plot([x*1e3 for x in a7[1]],[x/1e6 for x in a7[2]],'-k',linewidth=lw) ylabel(r'$\sigma_1$ [MPa]') #legend(('5k','10k','20k','40k'),2) #axis(ymin=0,ymax=10) ax2 = ax1.twinx() ax2.plot([x*1e3 for x in a1[1]],[x+y for x,y in zip(a1[5],a1[6])],'--b',linewidth=lw) ax2.plot([x*1e3 for x in a2[1]],[x+y for x,y in zip(a2[5],a2[6])],'--g',linewidth=lw) ax2.plot([x*1e3 for x in a3[1]],[x+y for x,y in zip(a3[5],a3[6])],'--r',linewidth=lw) ax2.plot([x*1e3 for x in a4[1]],[x+y for x,y in zip(a4[5],a4[6])],'--y',linewidth=lw) ax2.plot([x*1e3 for x in a5[1]],[x+y for x,y in zip(a5[5],a5[6])],'--m',linewidth=lw) ax2.plot([x*1e3 for x in a6[1]],[x+y for x,y in zip(a6[5],a6[6])],'--c',linewidth=lw) ax2.plot([x*1e3 for x in a7[1]],[x+y for x,y in zip(a7[5],a7[6])],'--k',linewidth=lw) ylabel('energy released by microcracking [J]') #axis(ymin=0,ymax=200) ### show or save show()

As you see in my code figure 4 appears as this: Floating numbers on the plot are too long

1.How can I change the precision of floating numbers on the plot? I would like to have only 3 decimal. Here is what I did:

A=[float("{0:.2e}".format(0.0023**0.5)), float("{0:.2e}".format(0.0019**0.5)), float("{0:.2e}".format(0.0015**0.5)), float("{0:.2e}".format(0.0012**0.5)), float("{0:.2e}".format(0.0009**0.5)), float("{0:.2e}".format(0.0008**0.5)), float("{0:.2e}".format(0.0007**0.5))] B=[float("{0:.2e}".format(max(a1[2])*((pi*hcl)**0.5))) ,float("{0:.2e}".format(max(a2[2])*((pi*hcl)**0.5))) ,float("{0:.2e}".format(max(a3[2])*((pi*hcl)**0.5))), float("{0:.2e}".format(max(a4[2])*((pi*hcl)**0.5))), float("{0:.2e}".format(max(a5[2])*((pi*hcl)**0.5))), float("{0:.2e}".format(max(a6[2])*((pi*hcl)**0.5))), float("{0:.2e}".format(max(a7[2])*((pi*hcl)**0.5)))]

It actually does the job but doesn't seem very intuitive to me. So the question is : Is there any other way to do the same thing? For instance how can I define the floating precision for my whole script?

Thanks

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  • Questions on SO are meant to be specific to one problem. You cannot post some code and ask to solve all the problems you have with it. If you have n different problems, ask n different questions, but don't use the same code, instead provide a minimal reproducible example for each individual problem. I've marked this question duplicate of 3 different questions, which are the 3 questions you're asking here. Commented May 28, 2017 at 9:37
  • @ImportanceOfBeingErnest , Thanks for your comment, I actually previously looked at those other questions you introduced. my case is quite different though. Thanks for the links anyway Commented May 28, 2017 at 11:33
  • @ImportanceOfBeingErnest, I edited my question and sort it in a more precise way. Hope this way is more clear. Commented May 28, 2017 at 11:35
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    My answer below also shows you how a minimal reproducible example should look like; because nobody cares about the subplots you have, the actual values, some data files or whatever else makes this 100 lines long, without being reproducible. Note that I did provide you with an answer now because this is your first question here, but in general if you don't stick to How to Ask, you will have problems getting your questions solved here. Commented May 28, 2017 at 11:58

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You don't want to change the floating point precision of the whole script.

What you want is only to format the labels with a certain precision. So you only want to set the format for the labels in the loop.

import numpy as np; np.random.seed(10) import matplotlib.pyplot as plt y = np.cumsum(np.random.normal(size=10)+0.2) x = np.arange(len(y)) fig, ax=plt.subplots() ax.plot(x,y) for a,b in zip(x,y): ax.annotate('({:d}, {:.2f})'.format(a,b), xy=(a,b), textcoords='data') plt.show()

enter image description here

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