创建一个白色居中于零的色彩映射
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：
-
这有效，但颜色相反
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将白色带走零点cs = plt.contourf(longrid, latgrid, ar[window-1]-bkgrd, levels, cmap = cm.get_cmap('BuRd_r', len(levels)-1))
我很感激任何帮助。
以下是测试色彩映射的功能和数据：
def PlotAnomalyCF(ar,hgrid,latgrid,longrid,outfile,levels,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,latgrid,
ar[window-1]-bkgrd,levels,
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,b,w,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, 0, 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. , 0.703125, 1.40625 , 2.109375,
2.8125 , 3.515625, 4.21875 , 4.921875, 5.625 ])
levels = array([-20, -19, -18, -17, -16, -15, -14, -13, -12, -11, -10, -9, -8,
-7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5,
6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18,
19, 20])
units ='$\mathrm{CC_{200}\,[\%]}$'
ar = shape is (13, 17, 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]])
1 个答案:
答案 0 :(得分：10)
我不会使用股票cmap，而是会完成你自己的制作。
正如您已经发现的那样，为了在使用cmaps时对颜色进行绝对控制（不通过颜色数组），cmap中的颜色数应该等于级别数 - 1。
我们可以通过以下示例轻松演示这一点。让我们在下面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, 0, 1),
(0, 1, 0),
(1, 0, 0)])
但是，当我们使用3 + 1级以外的任何其他值时，我们会得到您可能不会期望的结果：
plt.subplot(121)
plt.contourf(data, cmap=cmap, levels=[1, 4, 8, 10])
plt.subplot(122)
plt.contourf(data, cmap=cmap, levels=[1, 4, 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，所以我们也必须有偶数个等级。
好，现在我们有一些基本规则，让我们制作一个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, 0, 1),
(1, 1., 1),
(1, 0, 0)],
N=len(levs)-1,
)
plt.contourf(data, cmap=cmap, 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, 1.0)
希望这能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成功使用它。
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