Python 矩阵所有行对的相关系数和p值
我有一个矩阵Python 矩阵所有行对的相关系数和p值,python,numpy,statistics,scipy,correlation,Python,Numpy,Statistics,Scipy,Correlation,我有一个矩阵data,有m行和n列。我使用以下公式计算所有行对之间的相关系数: 现在我还想看看这些系数的p值np.corrcoef不提供这些;做但是,scipy.stats.pearsonr不接受输入矩阵 是否有快速方法计算所有行对的系数和p值(例如,到达两个m×m矩阵,一个具有相关系数,另一个具有相应的p值)无需手动检查所有对?最一致的方法可能是pandas中的buildin方法.corr,以获得r: In [79]: import pandas as pd m=np.random.rand
data
,有m行和n列。我使用以下公式计算所有行对之间的相关系数:
现在我还想看看这些系数的p值<代码>np.corrcoef不提供这些;做但是,scipy.stats.pearsonr
不接受输入矩阵
是否有快速方法计算所有行对的系数和p值(例如,到达两个m×m矩阵,一个具有相关系数,另一个具有相应的p值)无需手动检查所有对?最一致的方法可能是
pandas
中的buildin方法.corr
,以获得r:
In [79]:
import pandas as pd
m=np.random.random((6,6))
df=pd.DataFrame(m)
print df.corr()
0 1 2 3 4 5
0 1.000000 -0.282780 0.455210 -0.377936 -0.850840 0.190545
1 -0.282780 1.000000 -0.747979 -0.461637 0.270770 0.008815
2 0.455210 -0.747979 1.000000 -0.137078 -0.683991 0.557390
3 -0.377936 -0.461637 -0.137078 1.000000 0.511070 -0.801614
4 -0.850840 0.270770 -0.683991 0.511070 1.000000 -0.499247
5 0.190545 0.008815 0.557390 -0.801614 -0.499247 1.000000
要使用t-test获得p值:
In [84]:
n=6
r=df.corr()
t=r*np.sqrt((n-2)/(1-r*r))
import scipy.stats as ss
ss.t.cdf(t, n-2)
Out[84]:
array([[ 1. , 0.2935682 , 0.817826 , 0.23004382, 0.01585695,
0.64117917],
[ 0.2935682 , 1. , 0.04363408, 0.17836685, 0.69811422,
0.50661121],
[ 0.817826 , 0.04363408, 1. , 0.39783538, 0.06700715,
0.8747497 ],
[ 0.23004382, 0.17836685, 0.39783538, 1. , 0.84993082,
0.02756579],
[ 0.01585695, 0.69811422, 0.06700715, 0.84993082, 1. ,
0.15667393],
[ 0.64117917, 0.50661121, 0.8747497 , 0.02756579, 0.15667393,
1. ]])
In [85]:
ss.pearsonr(m[:,0], m[:,1])
Out[85]:
(-0.28277983892175751, 0.58713640696703184)
In [86]:
#be careful about the difference of 1-tail test and 2-tail test:
0.58713640696703184/2
Out[86]:
0.2935682034835159 #the value in ss.t.cdf(t, n-2) [0,1] cell
您也可以使用您在OP中提到的scipy.stats.pearsonr
:
In [95]:
#returns a list of tuples of (r, p, index1, index2)
import itertools
[ss.pearsonr(m[:,i],m[:,j])+(i, j) for i, j in itertools.product(range(n), range(n))]
Out[95]:
[(1.0, 0.0, 0, 0),
(-0.28277983892175751, 0.58713640696703184, 0, 1),
(0.45521036266021014, 0.36434799921123057, 0, 2),
(-0.3779357902414715, 0.46008763115463419, 0, 3),
(-0.85083961671703368, 0.031713908656676448, 0, 4),
(0.19054495489542525, 0.71764166168348287, 0, 5),
(-0.28277983892175751, 0.58713640696703184, 1, 0),
(1.0, 0.0, 1, 1),
#etc, etc
我今天遇到了同样的问题 在谷歌搜索了半个小时后,我在numpy/scipy库中找不到任何代码可以帮助我做到这一点 所以我写了自己版本的corrcoef
import numpy as np
from scipy.stats import pearsonr, betai
def corrcoef(matrix):
r = np.corrcoef(matrix)
rf = r[np.triu_indices(r.shape[0], 1)]
df = matrix.shape[1] - 2
ts = rf * rf * (df / (1 - rf * rf))
pf = betai(0.5 * df, 0.5, df / (df + ts))
p = np.zeros(shape=r.shape)
p[np.triu_indices(p.shape[0], 1)] = pf
p[np.tril_indices(p.shape[0], -1)] = p.T[np.tril_indices(p.shape[0], -1)]
p[np.diag_indices(p.shape[0])] = np.ones(p.shape[0])
return r, p
def corrcoef_loop(matrix):
rows, cols = matrix.shape[0], matrix.shape[1]
r = np.ones(shape=(rows, rows))
p = np.ones(shape=(rows, rows))
for i in range(rows):
for j in range(i+1, rows):
r_, p_ = pearsonr(matrix[i], matrix[j])
r[i, j] = r[j, i] = r_
p[i, j] = p[j, i] = p_
return r, p
第一个版本使用np.corrcoef的结果,然后根据corrcoef矩阵的三角形上限值计算p值
第二个循环版本只是迭代行,手动执行pearsonr
def test_corrcoef():
a = np.array([
[1, 2, 3, 4],
[1, 3, 1, 4],
[8, 3, 8, 5],
[2, 3, 2, 1]])
r1, p1 = corrcoef(a)
r2, p2 = corrcoef_loop(a)
assert np.allclose(r1, r2)
assert np.allclose(p1, p2)
测试通过了,他们是一样的
def test_timing():
import time
a = np.random.randn(100, 2500)
def timing(func, *args, **kwargs):
t0 = time.time()
loops = 10
for _ in range(loops):
func(*args, **kwargs)
print('{} takes {} seconds loops={}'.format(
func.__name__, time.time() - t0, loops))
timing(corrcoef, a)
timing(corrcoef_loop, a)
if __name__ == '__main__':
test_corrcoef()
test_timing()
我的Macbook在100x2500矩阵上的性能
corrcoef需要0.06608104705810547秒循环=10
corrcoef_循环需要7.585600137710571秒循环=10
有点粗俗,可能效率不高,但我认为这可能就是你想要的:
import scipy.spatial.distance as dist
import scipy.stats as ss
# Pearson's correlation coefficients
print dist.squareform(dist.pdist(data, lambda x, y: ss.pearsonr(x, y)[0]))
# p-values
print dist.squareform(dist.pdist(data, lambda x, y: ss.pearsonr(x, y)[1]))
是一个非常有用的函数,主要用于在n维空间中查找观测值之间的成对距离
但它允许用户定义的可调用“距离度量”,可以利用它执行任何类型的成对操作。结果以压缩距离矩阵形式返回,可以使用将其轻松更改为平方矩阵形式。如果不必使用,可以使用,因为它同时返回相关矩阵和p值(注意,前者要求数据为正态分布,而斯皮尔曼相关性为非参数度量,因此不假设数据为正态分布)。示例代码:
from scipy import stats
import numpy as np
data = np.array([[0, 1, -1], [0, -1, 1], [0, 1, -1]])
print 'np.corrcoef:', np.corrcoef(data)
cor, pval = stats.spearmanr(data.T)
print 'stats.spearmanr - cor:\n', cor
print 'stats.spearmanr - pval\n', pval
这与MATLAB中的corrcoef的性能完全相同: 要使此功能正常工作,您需要安装pandas和scipy
# Compute correlation correfficients matrix and p-value matrix
# Similar function as corrcoef in MATLAB
# dframe: pandas dataframe
def corrcoef(dframe):
fmatrix = dframe.values
rows, cols = fmatrix.shape
r = np.ones((cols, cols), dtype=float)
p = np.ones((cols, cols), dtype=float)
for i in range(cols):
for j in range(cols):
if i == j:
r_, p_ = 1., 1.
else:
r_, p_ = pearsonr(fmatrix[:,i], fmatrix[:,j])
r[j][i] = r_
p[j][i] = p_
return r, p
这里是@CT Zhu的答案的一个最小版本。我们不需要
pandas
,因为可以直接从numpy
计算相关性,这应该更快,因为我们不需要转换为数据帧的步骤
import numpy as np
import scipy.stats as ss
def corr_significance_two_sided(cc, nData):
# We will divide by 0 if correlation is exactly 1, but that is no problem
# We would simply set the test statistic to be infinity if it evaluates to NAN
with np.errstate(divide='ignore'):
t = -np.abs(cc) * np.sqrt((nData - 2) / (1 - cc**2))
t[t == np.nan] = np.inf
return ss.t.cdf(t, nData - 2) * 2 # multiply by two to get two-sided p-value
x = np.random.uniform(0, 1, (8, 1000))
cc = np.corrcoef(x)
pVal = corr_significance_two_sided(cc, 1000)
是否有理由不只是迭代行对?这有点笨拙,但代码不是很长,而且很可能不会出现性能问题,因为大部分时间都花在计算Pearson上。(也就是说,你是指编程时间上的“快”还是性能上的“快”)我建议您采用简单的方法并分析实际性能。您可以使用
metric='correlation'
,而不是通过自己的Python函数来计算相关系数,它等于(1-相关系数),并且用C编码(因此效率应该更高)。他也在寻找p值。如果你使用内置的相关度量,你将无法得到p值。你可以很容易地从相关系数中得出p值(参见jingchao的答案和)(也可以是CT Zhu的答案)这种方法满足了我的需要,对我来说似乎很简单。请遵循最适合您的答案。这段代码在scipy 1.0.0中失败,因为betai函数在弃用后已被删除。应该在scipy.special模块中使用betainc。感谢此解决方案,对我帮助很大!请注意,此i.0.0中的pValue比较相同功能时,实现设置为0(对角线上返回0)。但是,例如,scipy.stats.pearsonr
将返回这些情况下的p=1
。为了澄清,您的原始函数计算双边测试的p值,然后将其除以2以获得单边测试的p值,这是正确的吗?是的,在发布后的numpy和scipy中仍然没有实现这一点7年前
import numpy as np
import scipy.stats as ss
def corr_significance_two_sided(cc, nData):
# We will divide by 0 if correlation is exactly 1, but that is no problem
# We would simply set the test statistic to be infinity if it evaluates to NAN
with np.errstate(divide='ignore'):
t = -np.abs(cc) * np.sqrt((nData - 2) / (1 - cc**2))
t[t == np.nan] = np.inf
return ss.t.cdf(t, nData - 2) * 2 # multiply by two to get two-sided p-value
x = np.random.uniform(0, 1, (8, 1000))
cc = np.corrcoef(x)
pVal = corr_significance_two_sided(cc, 1000)