def plot_chi2(ax, LLR, DOF, yscale="log", c='b'):
ax.set_xlabel("LLR")
ax.set(yscale=yscale)
sns.distplot(LLR, color=c, kde=False, ax=ax, norm_hist=True, label="JMCTF")
q = np.linspace(0, np.max(LLR), 1000)
chi2 = tf.math.exp(tfd.Chi2(df=DOF).log_prob(q))
ax.plot(q, chi2, color=c, lw=2, label="chi^2 (DOF={0})".format(DOF))
def _init_distribution(conditions, **kwargs):
df = conditions["df"]
return tfd.Chi2(df=df, **kwargs)
#q_null, joint_fitted_null, all_pars_null = joint_null.fit_all(samples)
#print(all_pars_null)
#fig = plot_MLE_dist(all_pars_null)
#fig.tight_layout()
#fig.savefig("quickstart_MLE_dists.svg")
import matplotlib.pyplot as plt
import seaborn as sns
from jmctf.plotting import plot_sample_dist, plot_MLE_dist
from tensorflow_probability import distributions as tfd
fig = plt.figure(figsize=(5, 3))
ax = fig.add_subplot(111)
ax.set_xlabel("LLR")
ax.set(yscale="log")
sns.distplot(LLR, color='b', kde=False, ax=ax, norm_hist=True, label="JMCTF")
q = np.linspace(0, np.max(LLR), 1000)
chi2 = tf.math.exp(tfd.Chi2(df=DOF).log_prob(q))
ax.plot(q, chi2, color='b', lw=2, label="chi^2 (DOF={0})".format(DOF))
ax.legend(loc=1, frameon=False, framealpha=0, prop={'size': 10}, ncol=1)
fig.tight_layout()
fig.savefig("quickstart_LLR.svg")
# Or using helper tools:
fig = plt.figure(figsize=(5, 3))
ax = fig.add_subplot(111)
plot_chi2(ax, LLR, DOF)
ax.legend(loc=1, frameon=False, framealpha=0, prop={'size': 10}, ncol=1)
fig.tight_layout()
fig.savefig("quickstart_plot_LLR.svg")
color='b',
kde=False,
ax=ax1,
norm_hist=True,
label="LEEC bootstrap")
sns.distplot(bootstrap_chi2,
color='b',
kde=False,
ax=ax2,
norm_hist=True,
label="LEEC bootstrap")
qx = np.linspace(
0, np.max(chi2_quad),
1000) # 6 sigma too far for tf, cdf is 1. single-precision float I guess
qy = tf.math.exp(tfd.Chi2(df=DOF).log_prob(qx))
sns.lineplot(qx, qy, color='g', ax=ax1, label="asymptotic")
sns.lineplot(qx, qy, color='g', ax=ax2, label="asymptotic")
# Observed empirical and asymptotic p-values
epval = 1 - c.CDFf(chi2_quad)(chi2_quad_obs)
apval = tfd.Chi2(df=DOF).log_prob(chi2_quad_obs)
esig = -tfd.Normal(0, 1).quantile(epval)
asig = -tfd.Normal(0, 1).quantile(apval)
ax1.axvline(x=chi2_quad_obs,
lw=2,
c='k',
label="e_z={0}, a_z={1}".format(asig, esig))
ax2.axvline(x=chi2_quad_obs,
lw=2,
#MLLR = Lall(all_truepar) # Try without subtracting best fit
print("MLLR.shape:", MLLR.shape)
# Combined MLLR
LcombT_sep = LogLike(tf.expand_dims(xT, axis=0))
print("LcombT_sep:", LcombT_sep)
LcombT = tf.transpose(tf.reduce_sum(LcombT_sep, axis=1))
MLLR_comb = LcombT - Lcomb_bf[:, 0]
print("MLLR_comb.shape:", MLLR_comb.shape)
fig = plt.figure(figsize=(6, 4))
ax1 = fig.add_subplot(111)
ax1.set(yscale="log")
sns.distplot(MLLR[0], bins=50, kde=False, ax=ax1, norm_hist=True, label="MLLR")
q = np.linspace(np.min(MLLR[0]), np.max(MLLR[0]), 1000)
chi2 = tf.math.exp(tfd.Chi2(df=1).log_prob(q))
sns.lineplot(q, chi2, color='b', ax=ax1)
plt.tight_layout()
fig.savefig("ind_fits{0}.png".format(suffix))
# Now we need to select the signal region depending on the parameter values
# First, select based on the *test* point. This will always choose the
# same signal region, no matter the data, so it should work just fine.
truepar = tf.broadcast_to(xT, shape=(N, 1))
selected = tf.expand_dims(selector(truepar), 1)
print("selected:", selected)
ordinals = tf.reshape(tf.range(MLLR.shape[0]), (-1, 1))
idx = tf.stack([ordinals, selected], axis=-1)
MLLRsel = tf.gather_nd(MLLR, idx)
def _base_dist(self, nu: IntTensorLike, *args, **kwargs):
return tfd.Chi2(df=nu, *args, **kwargs)
framealpha=0,
prop={'size': 10},
ncol=1)
ax2.legend(loc=1,
frameon=False,
framealpha=0,
prop={'size': 10},
ncol=1)
if do_gof_tests:
#qx = np.linspace(0, tfd.Chi2(df=DOF).quantile(tfd.Normal(0,1).cdf(6.)),1000) # TODO: apparantly Quantile not implemented yet for Chi2 in tf
qx = np.linspace(
0,
sps.chi2(df=DOF).ppf(tfd.Normal(0, 1).cdf(5.)), 1000
) # 6 sigma too far for tf, cdf is 1. single-precision float I guess
qy = tf.math.exp(tfd.Chi2(df=DOF).log_prob(qx))
sns.lineplot(qx, qy, color='g', ax=ax3)
sns.lineplot(qx, qy, color='g', ax=ax4)
ax3.legend(loc=1,
frameon=False,
framealpha=0,
prop={'size': 10},
ncol=1)
ax4.legend(loc=1,
frameon=False,
framealpha=0,
prop={'size': 10},
ncol=1)
qx = np.linspace(
kde=False,
ax=ax2,
norm_hist=True,
label="LEEC quad")
sns.distplot(bootstrap_chi2,
color='b',
kde=False,
ax=ax1,
norm_hist=True,
label="LEEC bootstrap")
sns.distplot(bootstrap_chi2,
color='b',
kde=False,
ax=ax2,
norm_hist=True,
label="LEEC bootstrap")
qx = np.linspace(
0, np.max(chi2_quad),
1000) # 6 sigma too far for tf, cdf is 1. single-precision float I guess
qy = 0.5 * tf.math.exp(tfd.Chi2(df=DOF).log_prob(
qx)) # Half-chi2 since negative signal contributions not possible?
sns.lineplot(qx, qy, color='g', ax=ax1, label="asymptotic")
sns.lineplot(qx, qy, color='g', ax=ax2, label="asymptotic")
ax1.legend(loc=1, frameon=False, framealpha=0, prop={'size': 10}, ncol=1)
ax2.legend(loc=1, frameon=False, framealpha=0, prop={'size': 10}, ncol=1)
fig.tight_layout()
fig.savefig("{0}/LEEC_quad_{1}_{2}.png".format(path, master_name, nullname))
