我必须承认,了解如何创建或操作matplotlib的色彩图并不是一件容易的事.因此,我正在寻求一些帮助来解释和设置从蓝色(负)到红色(正)的色彩图,并且白色以零为中心.我想在contourf中使用这个cmap：
>这有效,但颜色是相反的
cs = plt.contourf(longrid,latgrid,ar[window-1]-bkGrd,levels,cmap = cm.get_cmap('BuRd',len(levels)-1))
>这里的问题是BuRd_r在零附近带走了白色
cs = plt.contourf(longrid,cmap = cm.get_cmap('BuRd_r',len(levels)-1))
我很感激任何帮助.
这是测试色彩映射的功能和数据：
def PlotAnomalyCF(ar,hgrid,longrid,outfile,units):
window = 1
tsize = 8
plt.close()
plt.figure(figsize=(11.7,4.3) )
plt.clf()
plt.cla()
bkGrd = bn.nanmean(ar[:],0)
for v in hgrid:OA
plt.subplot(1,len(hgrid),window)
plt.title(v,fontsize=tsize)
plt.subplots_adjust(left=0.07,bottom=0.75,right=0.98,top=0.92,wspace=0.12,hspace=0.98)
cs = plt.contourf(longrid,cmap=cm.get_cmap('BuRd_r',len(levels)-1))
cs.cmap.set_over('r')
plt.grid(True)
plt.axis('off')
window += 1
ax = plt.gca()
pos = ax.get_position()
l,b,w,h = pos.bounds
print l,h
cax = plt.axes([l-0.848,b-0.05,0.91,0.02]) # setup colorbar axes
cbar = plt.colorbar(cs,cax=cax,orientation='horizontal')
cl = plt.getp(cbar.ax,'xticklabels')
plt.setp(cl,fontsize=8)
# Add units text
plt.figtext(0.012,0.83,units,fontsize=tsize+1)
print 'outfile.',outfile
plt.savefig(outfile,dpi=900,orientation='landscape',format='pdf')
plt.show()
# Data
hgrid = array([-18,-15,-12,-9,-6,-3,3,6,9,12,15,18])
latgrid = array([ 5.61402391,4.91227095,4.21051798,3.508765,2.80701201,2.10525902,1.40350602,0.70175301,0.,-0.70175302,-1.40350604,-2.10525907,-2.8070121,-3.50876514,-4.21051818,-4.91227122,-5.61402427])
longrid = array([-5.625,-4.921875,-4.21875,-3.515625,-2.8125,-2.109375,-1.40625,-0.703125,0.703125,1.40625,2.109375,2.8125,3.515625,4.21875,4.921875,5.625 ])
levels = array([-20,-19,-18,-17,-16,-14,-13,-11,-10,-8,-7,-5,-4,-2,-1,1,2,4,5,7,8,10,11,13,14,16,17,18,19,20])
units ='$\mathrm{CC_{200}\,[\%]}$'
ar = shape is (13,17) with max = 82.4 and min = 45.5. It would be easier to just
generate some random data within these intervals. Alternatively just copy one of the
array:
ar[1] = array([[ 46.91224792,46.21374984,46.86130719,47.01021234,46.72626813,46.2288305,46.43835451,45.79325437,45.58271668,46.35872217,48.08725987,48.44553638,47.76519316,47.6366742,48.40425078,48.77756577,49.33566712],[ 46.83599932,46.84286989,47.33453309,46.55030976,46.80566458,46.53292035,47.02261763,47.41084421,47.38724565,47.91122826,49.21117552,49.45223641,49.97629913,50.44165439,51.08080398,50.79600723,49.968034 ],[ 47.42288313,47.07674124,46.7167639,46.11959218,46.95814111,46.88763807,47.79510368,48.50213272,49.14340301,50.0550682,50.96554707,51.70960776,52.76304827,53.13428506,53.01955687,52.57951586,51.91245273],[ 47.71067291,47.48154219,47.40131211,47.45929857,48.46118424,48.65199823,49.38156691,49.86137507,50.55394084,51.96604309,52.60579898,53.69096203,54.22750101,54.37757099,54.31517398,53.47697773,53.41809044],[ 48.20779565,48.58856851,48.75880829,49.40822878,50.03355014,51.44922083,52.00567831,52.99485667,53.69339127,53.58208129,53.88588998,55.24096208,55.24137628,55.38338399,55.30856415,55.0329081,54.58041914],[ 49.20063728,49.81223264,50.2145489,50.54749112,51.44252761,53.1708726,54.48141824,55.1337493,55.86338227,55.80719304,56.1060897,56.15050406,56.10404113,56.82550383,57.12370494,56.79250814,57.21656741],[ 50.17222332,50.78494993,51.47036476,51.78513471,54.07329312,55.12136894,56.63202678,56.77587861,57.60688855,57.31874243,57.86532727,58.38753463,58.52204736,58.8451274,59.17282185,58.93137673,58.90977463],[ 51.50642331,51.51055372,52.44746806,53.37696513,54.86775802,56.68992167,57.90624404,58.76394172,59.64662899,59.80540837,59.98355254,60.05761821,59.95848562,60.54540623,60.4776266,60.11749116,59.74209418],[ 51.40396463,51.48043239,52.89530187,53.73500868,55.39612502,56.70178532,58.07064267,59.56644298,60.47288049,60.59095081,61.26474813,61.20278944,61.43807574,61.1942828,60.40014922,59.78371327,59.50410992],[ 51.89656984,52.18725649,52.57764233,54.39502415,55.61672911,57.04180061,58.54357871,59.76354498,60.24155861,60.59473182,60.65985503,60.92762915,60.76726905,60.47166256,60.42044548,60.04043031,60.06031171],[ 51.69698671,52.39494994,52.71685017,53.65488505,54.52480831,56.33091376,57.87811829,58.36719736,59.31479758,59.61329074,59.81807224,59.44053305,59.25522337,59.5309563,59.68850776,59.54046914,58.60604327],[ 50.81523448,51.5690217,51.81382216,52.88514118,53.27394887,54.6941369,55.81938471,56.85401174,57.10874072,58.55074569,58.45869901,57.67496274,57.3895956,58.05319653,58.83009123,57.90678384,56.97717781],[ 49.61355365,49.92298625,50.56281781,51.22117266,51.98020374,53.0328832,53.76602714,54.94396421,55.16849545,55.83033544,56.19720112,56.9382048,57.20669361,56.76351885,56.93632305,56.16428665,54.7570241 ],[ 49.17250426,49.14663096,49.86504335,50.24394193,50.84671744,51.21785439,51.72667942,52.8287256,53.65550669,53.81801262,54.62490542,54.88045696,55.27367041,54.89234901,54.51005786,53.57284022,52.53966996],[ 48.55125023,48.99467099,49.80237362,49.67943854,49.45569362,49.3387854,50.129718,49.94784906,51.03357894,51.73021167,52.0211617,52.55805334,52.22956095,51.88844202,50.87410618,50.39506101,49.72117909],[ 48.37551433,48.30812683,48.45300884,48.50818535,47.76798967,47.21965588,47.60764424,48.21327283,48.52370448,49.95221655,49.7608777,50.1178807,50.15020093,49.4369175,49.16839811,49.31254753,48.03208233],[ 46.91675497,46.45280928,46.37122283,46.69881125,45.95493853,46.33296801,46.15091804,46.09862998,46.31676066,46.6199912,47.60040926,48.34096053,47.78005438,48.0951173,48.15291404,47.21140107,46.28884057]])
解决方法
我将不再使用股票cmap,而是完成自己的生产.
正如您已经发现的那样,为了在使用cmaps时绝对控制颜色(不通过颜色数组),cmap中的颜色数应该等于级数 – 1.
我们可以通过以下示例轻松演示这一点.让我们用三种基色制作一个色彩图：
import matplotlib.pyplot as plt import matplotlib.colors as mcolors import numpy data = numpy.arange(12).reshape(3,4) cmap = mcolors.ListedColormap([(0,1),(0,0),(1,0)])
但是,当我们使用3 3级以外的任何其他值时,我们将得到您可能不会期望的结果：
plt.subplot(121) plt.contourf(data,cmap=cmap,levels=[1,10]) plt.subplot(122) plt.contourf(data,8]) plt.show()
所以我已经证明你的等级必须是n_colors-1的长度,但是因为我们想要有3种颜色,并且总是显示中间颜色,所以我们还必须具有偶数个等级.
好,现在我们有一些基本规则,让我们创建一个LinearSegmentedColormap,它实际上只是一个ListedColormap,插值为N种颜色：
levs = range(12) assert len(levs) % 2 == 0,'N levels must be even.' cmap = mcolors.LinearSegmentedColormap.from_list(name='red_white_blue',colors =[(0,1.,0)],N=len(levs)-1,) plt.contourf(data,levels=levs) plt.show()
我们可以通过编程验证normailzed cmap是否为白色作为中间颜色：
>>> print cmap(0.5) (1.0,1.0,1.0)
希望这将为您提供所需的所有信息,以便您能够制作任何您喜欢的色彩图并在轮廓上成功使用它.