matplot pie: rotate labels horizontally

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你的背包 2021-01-06 12:03

Using matplot to create a little gauge with the code below:

    group_size=[10,10,10,10,10,50]
    labels=[\'AAAA\',\'BBBB\',\'CCCC\',\'DDDD\',\'EEEE\',\'\'         


        
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  • 2021-01-06 12:19

    You need to rotate the pie labels manually. To this end you may loop over the labels and set the rotation to your needs.

    group_size=[10,10,10,10,10,50]
    labels=['AAAA','BBBB','CCCC','DDDD','EEEE','']
    fig, ax = plt.subplots()
    ax.axis('equal')
    pie = ax.pie(group_size, radius=2.2, colors=['k'] ,startangle=180,counterclock=False)
    pie2 = ax.pie([10,10,10,10,10,50], radius=2,  labeldistance=0.9, labels=labels, 
                  rotatelabels =True, startangle=180,counterclock=False)
    
    plt.setp(pie2[1], rotation_mode="anchor", ha="center", va="center")
    for tx in pie2[1]:
        rot = tx.get_rotation()
        tx.set_rotation(rot+90+(1-rot//180)*180)
    
    plt.show()
    

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  • 2021-01-06 12:34

    My solution is basically similar to the accepted answer by ImportanceOfBeingErnest. But I think the steps involved are easier to grasp.

    import matplotlib.pyplot as plt
    
    group_size = [10, 10, 10, 10, 10, 50]
    labels = ['AAAA', 'BBBB', 'CCCC', 'DDDD', 'EEEE', '']
    fig, ax = plt.subplots()
    ax.axis('equal')
    pie = ax.pie(group_size, radius=2.2, colors=['k'], startangle=180, counterclock=False)
    
    # ax.pie() returns wedges, labels
    # Note that rotatelabels=False; so that, at this step ...
    # all labels are not rotated; they will be rotated later
    wedges, labels = ax.pie([10, 10, 10, 10, 10, 50], radius=2,  \
                          labeldistance=0.85, labels=labels, rotatelabels = False, \
                          startangle=180, counterclock=False)
    
    # do the rotation of the labels
    for ea, eb in zip(wedges, labels):
        mang =(ea.theta1 + ea.theta2)/2.  # get mean_angle of the wedge
        #print(mang, eb.get_rotation())
        eb.set_rotation(mang+270)         # rotate the label by (mean_angle + 270)
        eb.set_va("center")
        eb.set_ha("center")
    
    plt.show()
    

    The resulting image:

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