如何旋轉極坐標matplotlib圖中的刻度標簽？

Question

我有很長的標識符，我想做一個徑向圖，刻度線都在不同的角度。 例如，右側0度處的第一個刻度應具有0度角。 頂部的角度應為90度。 左側270度的角度應為0度。 我希望它看起來像是徑向樹狀圖。 使用matplotlib 2.0.2和python 3.6.2

在matplotlib中是否可以旋轉單獨的刻度標簽或單獨添加文本標簽？

注意：我已經更新了繪圖，以響應以下@ImportanceOfBeingErnest。

設置ax.set_rticks([])會在添加散點和線時使圖變形。 label.get_position()的位置使標簽大大偏離了繪圖的右側。

有沒有一種使用角度和幅度坐標的方法？

import pandas as pd
import matplotlib.pyplot as plt
import numpy as np

data = {'0-b__|ce0Ji|aaaiIi9abGc_|ti5l-baa1tcciii|irGi': 0.28774963897009614, '0-b__|ce0Ji|aaaiIi9abGc_|ti6l-baa1tcciii|irGi': 0.18366735937444964, 'allb_e__|tla1a|ali|_auc7en_|e': -0.11720263463773731, 'b__0|lp|..ii80p.e7l_|an4obln.llll0ai|': -0.021168680215561269, 'b__Ass8._ii8.c4on|Ay|mbessoxiAxa': 0.17845443978725653, 'b__Bts4o_rrtiordae|Bei|obe7rattrniBno': 0.32077066676059313, 'b__|aaa|tteiatlim_|e1rblttaaeei|e': -0.27915536613715614, 'b__|as4.|ei2.l7ov_|e0tblaaoxi|xa': 0.43309499489274772, 'b__|as4.|ei2.l7ov_|e9tblaaoxi|xa': 0.47835581698425556, 'b__|cu|ppripcae_|co2tbopnccpei|': -0.20330386390053184, 'b__|eoea|cccimacnuuh_|ra0obarceenbi|ba': 0.062889648127927869, 'b__|oa|ggrigoip_|nr6ybmgvoohii|i': -0.045648268817583035, 'b__|p1|ooiioi4rs_|sr5eba0otsoi|ox': -0.52544820541720971, 'b__|paa|piatgn_|hy1cboippoli|la': 0.27260399422352155, 'b__|triu|mmriumay_|eb4ebcimrttnhi|hc': 0.62680074671550845, 'b__|tru|mmriumad_|eb2obcmittisi|': 0.34780388151174668, 'etob_m__|aol2l|ooeui|_lool7r': 0.4856468599203973, 'etpb_s__|apl2l|lleni|_loll8e': 0.24430277200521291, 'ib__rCalc_hhdiorchubai|CSt|absahodrsiCsaaca': -0.13484907188897891, 'nlab___|oa1i|ssni|_iesa9': 0.13636363636363635, 'nlnb_i__|dn1t|rrnfi|_tera8ig_|e': -0.056954668733049205, 'nrfb_h__|afl3r|ssnti|_resl3yn_': 0.56102285935683849, 'o5b__l|rcoa|eecialaeprh_|as1o5bie0trrnlii|irLa': 0.53377831002782572, 'oelb_a__Aelt3_rrovi__rro|a': 0.32230284245007218, 'oelb_a__Aelt4_rrovi__rro|a': 0.16580958754534889, 'porb_i__Ctrc6c_oopci__cloa|ny|C': 0.38260364199922509, 'porb_i__Ctrc7g_rrpci__glra|ay|C': 0.51829805219964076, 'ptab_a__|hac2b|uupci|_boui3ct_|': 0.50873516255151285, 'reab_a__|aa2a|rrrhi|_axrl4ra_|': -0.47742242259871087, 'sb__o|sSac|ccnibocsctlhd_|a0dbuacmssioai|anCca': 0.42733612764608503, 'teob___|oa1b|iiti|_bnil3': -0.32684653587404461, 'uoib_i__|ia2a|bbuli|_arbi2it': -0.13636363636363635}
Se_corr = pd.Series(data, name="correlation")


def plot_polar(r):
    with plt.style.context("seaborn-whitegrid"):
        fig = plt.figure(figsize=(10,10))
        ax = fig.add_subplot(111, polar=True)
        ax.set_rmax(2)
#         ax.set_rticks([])
        ticks= np.linspace(0, 360, r.index.size + 1) [:-1]

        ax.set_xticks(np.deg2rad(ticks))
        ax.set_xticklabels(r.index, fontsize=15,)

        angles = np.linspace(0,2*np.pi,len(ax.get_xticklabels()))
        angles[np.cos(angles) < 0] = angles[np.cos(angles) < 0] + np.pi
        angles = np.rad2deg(angles)

        for i, theta in enumerate(angles):
            ax.plot([theta,theta], [0,r[i]], color="black")
            ax.scatter(x=theta,y=r[i], color="black")


        labels = []
        for label, theta in zip(ax.get_xticklabels(), angles):
            x,y = label.get_position()
            lab = ax.text(x, y, label.get_text(), transform=label.get_transform(),
                          ha=label.get_ha(), va=label.get_va())
            lab.set_rotation(theta)
            labels.append(lab)
        ax.set_xticklabels([])

    return fig, ax 
fig,ax = plot_polar(Se_corr)

Answer 1

旋轉極坐標圖的刻度標簽可能不像通常的笛卡爾圖那樣容易。 對於笛卡爾圖，可以簡單地做

for label in ax.get_xticklabels():
    label.set_rotation(...)

這不適用於極坐標圖，因為在繪制時將其旋轉重置為0度。

我想到的一種選擇是創建新的刻度標簽作為其他文本對象，這些對象復制刻度標簽的屬性，但可以持續旋轉。 然后刪除所有原始的滴答標簽。

import numpy as np
import matplotlib.pyplot as plt

r = np.arange(0, 2, 0.01)
theta = 2 * np.pi * r

ax = plt.subplot(111, projection='polar')
ax.plot(theta, r)
ax.set_rmax(2)
ax.set_rticks([]) 


plt.gcf().canvas.draw()
angles = np.linspace(0,2*np.pi,len(ax.get_xticklabels())+1)
angles[np.cos(angles) < 0] = angles[np.cos(angles) < 0] + np.pi
angles = np.rad2deg(angles)
labels = []
for label, angle in zip(ax.get_xticklabels(), angles):
    x,y = label.get_position()
    lab = ax.text(x,y, label.get_text(), transform=label.get_transform(),
                  ha=label.get_ha(), va=label.get_va())
    lab.set_rotation(angle)
    labels.append(lab)
ax.set_xticklabels([])

plt.show()

對於更長的標簽，您可以使用標簽的y坐標：

import numpy as np
import matplotlib.pyplot as plt

r = np.arange(0, 2, 0.01)
theta = 2 * np.pi * r

ax = plt.subplot(111, projection='polar')
ax.plot(theta, r)
ax.set_rmax(2)
ax.set_rticks([])
ticks= np.linspace(0,360,9)[:-1] 
ax.set_xticks(np.deg2rad(ticks))
ticklabels = ["".join(np.random.choice(list("ABCDE"),size=15)) for _ in range(len(ticks))]
ax.set_xticklabels(ticklabels, fontsize=10)

plt.gcf().canvas.draw()
angles = np.linspace(0,2*np.pi,len(ax.get_xticklabels())+1)
angles[np.cos(angles) < 0] = angles[np.cos(angles) < 0] + np.pi
angles = np.rad2deg(angles)
labels = []
for label, angle in zip(ax.get_xticklabels(), angles):
    x,y = label.get_position()
    lab = ax.text(x,y-.65, label.get_text(), transform=label.get_transform(),
                  ha=label.get_ha(), va=label.get_va())
    lab.set_rotation(angle)
    labels.append(lab)
ax.set_xticklabels([])

plt.subplots_adjust(top=0.68,bottom=0.32,left=0.05,right=0.95)
plt.show()

---

修改后的問題代碼的更正版本：

 import pandas as pd import matplotlib.pyplot as plt import numpy as np data = {'0-b__|ce0Ji|aaaiIi9abGc_|ti5l-baa1tcciii|irGi': 0.28774963897009614, '0-b__|ce0Ji|aaaiIi9abGc_|ti6l-baa1tcciii|irGi': 0.18366735937444964, 'allb_e__|tla1a|ali|_auc7en_|e': -0.11720263463773731, 'b__0|lp|..ii80p.e7l_|an4obln.llll0ai|': -0.021168680215561269, 'b__Ass8._ii8.c4on|Ay|mbessoxiAxa': 0.17845443978725653, 'b__Bts4o_rrtiordae|Bei|obe7rattrniBno': 0.32077066676059313, 'b__|aaa|tteiatlim_|e1rblttaaeei|e': -0.27915536613715614, 'b__|as4.|ei2.l7ov_|e0tblaaoxi|xa': 0.43309499489274772, 'b__|as4.|ei2.l7ov_|e9tblaaoxi|xa': 0.47835581698425556, 'b__|cu|ppripcae_|co2tbopnccpei|': -0.20330386390053184, 'b__|eoea|cccimacnuuh_|ra0obarceenbi|ba': 0.062889648127927869, 'b__|oa|ggrigoip_|nr6ybmgvoohii|i': -0.045648268817583035, 'b__|p1|ooiioi4rs_|sr5eba0otsoi|ox': -0.52544820541720971, 'b__|paa|piatgn_|hy1cboippoli|la': 0.27260399422352155, 'b__|triu|mmriumay_|eb4ebcimrttnhi|hc': 0.62680074671550845, 'b__|tru|mmriumad_|eb2obcmittisi|': 0.34780388151174668, 'etob_m__|aol2l|ooeui|_lool7r': 0.4856468599203973, 'etpb_s__|apl2l|lleni|_loll8e': 0.24430277200521291, 'ib__rCalc_hhdiorchubai|CSt|absahodrsiCsaaca': -0.13484907188897891, 'nlab___|oa1i|ssni|_iesa9': 0.13636363636363635, 'nlnb_i__|dn1t|rrnfi|_tera8ig_|e': -0.056954668733049205, 'nrfb_h__|afl3r|ssnti|_resl3yn_': 0.56102285935683849, 'o5b__l|rcoa|eecialaeprh_|as1o5bie0trrnlii|irLa': 0.53377831002782572, 'oelb_a__Aelt3_rrovi__rro|a': 0.32230284245007218, 'oelb_a__Aelt4_rrovi__rro|a': 0.16580958754534889, 'porb_i__Ctrc6c_oopci__cloa|ny|C': 0.38260364199922509, 'porb_i__Ctrc7g_rrpci__glra|ay|C': 0.51829805219964076, 'ptab_a__|hac2b|uupci|_boui3ct_|': 0.50873516255151285, 'reab_a__|aa2a|rrrhi|_axrl4ra_|': -0.47742242259871087, 'sb__o|sSac|ccnibocsctlhd_|a0dbuacmssioai|anCca': 0.42733612764608503, 'teob___|oa1b|iiti|_bnil3': -0.32684653587404461, 'uoib_i__|ia2a|bbuli|_arbi2it': -0.13636363636363635} Se_corr = pd.Series(data, name="correlation") def plot_polar(r): with plt.style.context("seaborn-whitegrid"): fig = plt.figure(figsize=(10,10)) ax = fig.add_subplot(111, polar=True) ax.set_rmax(2) #ax.set_rticks([]) ticks= np.linspace(0, 360, r.index.size + 1)[:-1] ax.set_xticks(np.deg2rad(ticks)) ax.set_xticklabels(r.index, fontsize=15,) angles = np.linspace(0,2*np.pi,len(ax.get_xticklabels())+1) angles[np.cos(angles) < 0] = angles[np.cos(angles) < 0] + np.pi angles = np.rad2deg(angles) for i, theta in enumerate(angles[:-1]): ax.plot([theta,theta], [0,r[i]], color="black") ax.scatter(x=theta,y=r[i], color="black") fig.canvas.draw() labels = [] for label, theta in zip(ax.get_xticklabels(), angles): x,y = label.get_position() lab = ax.text(x, y, label.get_text(), transform=label.get_transform(), ha=label.get_ha(), va=label.get_va()) lab.set_rotation(theta) labels.append(lab) ax.set_xticklabels([]) return fig, ax fig,ax = plot_polar(Se_corr) plt.show() 

; 該代碼產生的圖像