应用来自熊猫df
我正在尝试plot multivariate distribution产生的multiple xy。
下面的coordinates旨在获取每个坐标并将其应用为半径（code）。然后通过[_Rad]因子（COV）调整matrix scaling，以扩展[_Scaling]中的半径并收缩x-direction中的半径。其方向由y-direction rotation（angle）度量。
输出表示为[_Rotation]函数，该函数表示每个组坐标在特定空间上的影响。
尽管，目前我只能将probability应用于code中的最后一组coordinates。因此，使用下面的输入，只有df有效。 A3_X, A3_Y和A1_X, A1_Y, A2_X, A2_Y。请参阅附图以直观表示。
注意：很长的B1_X, B1_Y, B2_X, B2_Y表示歉意。这是复制我的df的唯一方法。
dataset如下所示。 import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts
def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])
def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])
r = rot(theta)
return r @ cov @ r.T
def datalimits(*data, pad=.15):
dmin,dmax = min(d.values.min() for d in data), max(d.values.max() for d in data)
spad = pad*(dmax - dmin)
return dmin - spad, dmax + spad
d = ({
'Time' : [1],
'A1_Y' : [5883.102906],
'A1_X' : [3321.527705],
'A2_Y' : [5898.467202],
'A2_X' : [3328.331657],
'A3_Y' : [5886.270552],
'A3_X' : [3366.777169],
'B1_Y' : [5897.925245],
'B1_X' : [3297.143092],
'B2_Y' : [5905.137781],
'B2_X' : [3321.167842],
'B3_Y' : [5888.291025],
'B3_X' : [3347.263205],
'A1_Radius' : [10.3375199],
'A2_Radius' : [10.0171423],
'A3_Radius' : [11.42129333],
'B1_Radius' : [18.69514267],
'B2_Radius' : [10.65877044],
'B3_Radius' : [9.947025444],
'A1_Scaling' : [0.0716513620],
'A2_Scaling' : [0.0056262380],
'A3_Scaling' : [0.0677243260,],
'B1_Scaling' : [0.0364290850],
'B2_Scaling' : [0.0585827450],
'B3_Scaling' : [0.0432806750],
'A1_Rotation' : [20.58078926],
'A2_Rotation' : [173.5056346],
'A3_Rotation' : [36.23648405],
'B1_Rotation' : [79.81849817],
'B2_Rotation' : [132.2437404],
'B3_Rotation' : [44.28198078],
})
df = pd.DataFrame(data=d)
A_Y = df[df.columns[1::2][:3]]
A_X = df[df.columns[2::2][:3]]
B_Y = df[df.columns[7::2][:3]]
B_X = df[df.columns[8::2][:3]]
A_Radius = df[df.columns[13:16]]
B_Radius = df[df.columns[16:19]]
A_Scaling = df[df.columns[19:22]]
B_Scaling = df[df.columns[22:25]]
A_Rotation = df[df.columns[25:28]]
B_Rotation = df[df.columns[28:31]]
limitpad = .5
clevels = 5
cflevels = 50
xmin,xmax = datalimits(A_X, B_X, pad=limitpad)
ymin,ymax = datalimits(A_Y, B_Y, pad=limitpad)
X,Y = np.meshgrid(np.linspace(xmin, xmax), np.linspace(ymin, ymax))
fig = plt.figure(figsize=(10,6))
ax = plt.gca()
Zs = []
for l,color in zip('AB', ('red', 'blue')):
ax.plot(A_X.iloc[0], A_Y.iloc[0], '.', c='red', ms=10, label=l, alpha = 0.6)
ax.plot(B_X.iloc[0], B_Y.iloc[0], '.', c='blue', ms=10, label=l, alpha = 0.6)
Zrows = []
for _,row in df.iterrows():
for i in [1,2,3]:
x,y = row['{}{}_X'.format(l,i)], row['{}{}_Y'.format(l,i)]
cov = getcov(radius=row['{}{}_Radius'.format(l,i)],scale=row['{}{}_Scaling'.format(l,i)], theta=row['{}{}_Rotation'.format(l,i)])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))
Zrows.append(Z)
Zs.append(np.sum(Zrows, axis=0))
Z = Zs[0] - Zs[1]
normZ = Z - Z.min()
normZ = normZ/normZ.max()
cs = ax.contour(X, Y, normZ, levels=clevels, colors='w', alpha=.5)
ax.clabel(cs, fmt='%2.1f', colors='w')#, fontsize=14)
cfs = ax.contourf(X, Y, normZ, levels=cflevels, cmap='viridis', vmin=0, vmax=1)
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])
仅适用于code和A3_X, A3_Y。
它不适用于坐标B3_X, B3_Y和A1_X, A1_Y, A2_X, A2_Y。
3 个答案:
答案 0 :(得分：1)
尤其是在中间嵌套的for循环中，只需调整缩进即可，并在遍历数据帧行时重置 Zrows 列表。查看代码中的注释以了解具体更改：
...
for _, row in df.iterrows():
# MOVE ZROWS INSIDE
Zrows = []
for i in [1,2,3]:
x,y = row['{}{}_X'.format(l,i)], row['{}{}_Y'.format(l,i)]
# INDENT cov AND LATER CALCS TO RUN ACROSS ALL 1,2,3
cov = getcov(radius=row['{}{}_Radius'.format(l,i)],
scale=row['{}{}_Scaling'.format(l,i)],
theta=row['{}{}_Rotation'.format(l,i)])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))
# APPEND TO BE CLEANED OUT WITH EACH ROW
Zrows.append(Z)
Zs.append(np.sum(Zrows, axis=0))
...
答案 1 :(得分：1)
迭代点数据的方式有误。数据框的组织方式使得很难找出适当的方法来遍历数据，并且很容易遇到所遇到的错误。如果将df的组织方式设置得更好，那么您就可以轻松地遍历代表每个组A和B的数据子集。如果您从数据字典d中分配时间，则可以通过以下方法简化使用df的工作：
import pandas as pd
time = [1]
d = ({
'A1_Y' : [5883.102906],
'A1_X' : [3321.527705],
'A2_Y' : [5898.467202],
'A2_X' : [3328.331657],
'A3_Y' : [5886.270552],
'A3_X' : [3366.777169],
'B1_Y' : [5897.925245],
'B1_X' : [3297.143092],
'B2_Y' : [5905.137781],
'B2_X' : [3321.167842],
'B3_Y' : [5888.291025],
'B3_X' : [3347.263205],
'A1_Radius' : [10.3375199],
'A2_Radius' : [10.0171423],
'A3_Radius' : [11.42129333],
'B1_Radius' : [18.69514267],
'B2_Radius' : [10.65877044],
'B3_Radius' : [9.947025444],
'A1_Scaling' : [0.0716513620],
'A2_Scaling' : [0.0056262380],
'A3_Scaling' : [0.0677243260,],
'B1_Scaling' : [0.0364290850],
'B2_Scaling' : [0.0585827450],
'B3_Scaling' : [0.0432806750],
'A1_Rotation' : [20.58078926],
'A2_Rotation' : [173.5056346],
'A3_Rotation' : [36.23648405],
'B1_Rotation' : [79.81849817],
'B2_Rotation' : [132.2437404],
'B3_Rotation' : [44.28198078],
})
# a list of tuples of the form ((time, group_id, point_id, value_label), value)
tuples = [((t, k.split('_')[0][0], int(k.split('_')[0][1]), k.split('_')[1]), v[i]) for k,v in d.items() for i,t in enumerate(time)]
df = pd.Series(dict(tuples)).unstack(-1)
df.index.names = ['time', 'group', 'id']
print(df)
输出：
Radius Rotation Scaling X Y
time group id
1 A 1 10.337520 20.580789 0.071651 3321.527705 5883.102906
2 10.017142 173.505635 0.005626 3328.331657 5898.467202
3 11.421293 36.236484 0.067724 3366.777169 5886.270552
B 1 18.695143 79.818498 0.036429 3297.143092 5897.925245
2 10.658770 132.243740 0.058583 3321.167842 5905.137781
3 9.947025 44.281981 0.043281 3347.263205 5888.291025
这将使遍历数据中的子集变得更加容易。以下是您在每个时间点遍历每个组的子数据帧的方法：
for time, tdf in df.groupby('time'):
for group, gdf in tdf.groupby('group'):
...
这是my code from your previous question的更新版本，它使用组织得更好的数据框在每个时间点创建所需的绘图：
for time,subdf in df.groupby('time'):
plotmvs(subdf)
输出：
以下是上述plotmvs函数的完整代码：
import numpy as np
import pandas as pd
from mpl_toolkits.axes_grid1 import make_axes_locatable
import matplotlib.pyplot as plt
import scipy.stats as sts
def datalimits(*data, pad=.15):
dmin,dmax = min(d.min() for d in data), max(d.max() for d in data)
spad = pad*(dmax - dmin)
return dmin - spad, dmax + spad
def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])
def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])
r = rot(theta)
return r @ cov @ r.T
def mvpdf(x, y, xlim, ylim, radius=1, velocity=0, scale=0, theta=0):
"""Creates a grid of data that represents the PDF of a multivariate gaussian.
x, y: The center of the returned PDF
(xy)lim: The extent of the returned PDF
radius: The PDF will be dilated by this factor
scale: The PDF be stretched by a factor of (scale + 1) in the x direction, and squashed by a factor of 1/(scale + 1) in the y direction
theta: The PDF will be rotated by this many degrees
returns: X, Y, PDF. X and Y hold the coordinates of the PDF.
"""
# create the coordinate grids
X,Y = np.meshgrid(np.linspace(*xlim), np.linspace(*ylim))
# stack them into the format expected by the multivariate pdf
XY = np.stack([X, Y], 2)
# displace xy by half the velocity
x,y = rot(theta) @ (velocity/2, 0) + (x, y)
# get the covariance matrix with the appropriate transforms
cov = getcov(radius=radius, scale=scale, theta=theta)
# generate the data grid that represents the PDF
PDF = sts.multivariate_normal([x, y], cov).pdf(XY)
return X, Y, PDF
def mvpdfs(xs, ys, xlim, ylim, radius=None, velocity=None, scale=None, theta=None):
PDFs = []
for i,(x,y) in enumerate(zip(xs,ys)):
kwargs = {
'radius': radius[i] if radius is not None else 1,
'velocity': velocity[i] if velocity is not None else 0,
'scale': scale[i] if scale is not None else 0,
'theta': theta[i] if theta is not None else 0,
'xlim': xlim,
'ylim': ylim
}
X, Y, PDF = mvpdf(x, y, **kwargs)
PDFs.append(PDF)
return X, Y, np.sum(PDFs, axis=0)
def plotmvs(df, xlim=None, ylim=None, fig=None, ax=None):
"""Plot an xy point with an appropriately tranformed 2D gaussian around it.
Also plots other related data like the reference point.
"""
if xlim is None: xlim = datalimits(df['X'])
if ylim is None: ylim = datalimits(df['Y'])
if fig is None:
fig = plt.figure(figsize=(8,8))
ax = fig.gca()
elif ax is None:
ax = fig.gca()
PDFs = []
for (group,gdf),color in zip(df.groupby('group'), ('red', 'blue')):
# plot the xy points of each group
ax.plot(*gdf[['X','Y']].values.T, '.', c=color)
# fetch the PDFs of the 2D gaussian for each group
kwargs = {
'radius': gdf['Radius'].values if 'Radius' in gdf else None,
'velocity': gdf['Velocity'].values if 'Velocity' in gdf else None,
'scale': gdf['Scaling'].values if 'Scaling' in gdf else None,
'theta': gdf['Rotation'].values if 'Rotation' in gdf else None,
'xlim': xlim,
'ylim': ylim
}
X, Y, PDF = mvpdfs(gdf['X'].values, gdf['Y'].values, **kwargs)
PDFs.append(PDF)
# create the PDF for all points from the difference of the sums of the 2D Gaussians from group A and group B
PDF = PDFs[0] - PDFs[1]
# normalize PDF by shifting and scaling, so that the smallest value is 0 and the largest is 1
normPDF = PDF - PDF.min()
normPDF = normPDF/normPDF.max()
# plot and label the contour lines of the 2D gaussian
cs = ax.contour(X, Y, normPDF, levels=6, colors='w', alpha=.5)
ax.clabel(cs, fmt='%.3f', fontsize=12)
# plot the filled contours of the 2D gaussian. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normPDF, levels=50, cmap='viridis')
# create the colorbar and ensure that it goes from 0 -> 1
divider = make_axes_locatable(ax)
cax = divider.append_axes("right", size="5%", pad=0.1)
cbar = fig.colorbar(cfs, ax=ax, cax=cax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])
# ensure that x vs y scaling doesn't disrupt the transforms applied to the 2D gaussian
ax.set_aspect('equal', 'box')
return fig, ax
答案 2 :(得分：0)
这段代码中发生了很多事情。我注意到的一件事是，您似乎没有正确使用df.columns索引。如果您查看A_Y，则输出为：
A1_Rotation A1_X A2_Radius
0 20.580789 3321.527705 10.017142
我认为您正在混合列。也许使用df[['A1_Y', 'A2_Y', 'A3_Y']]来获取确切的列，或者只是将所有A_Y值放入单个列中。
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