获取nlme.lme()或lme4.lmer()在RPy中的简洁总结
我正在通过RPy与nlme和lme4这两个R函数进行交互，我想在我的Python控制台中获取一个输出摘要。
我运行了以下代码：
test1=nlme.lme(r.formula('Pupil~CoI*Time'), random=r.formula('~1|ID'),data=dfr)
test2=nlme.lme(r.formula('Pupil~CoI*measurement'),random=r.formula('~1|ID'),data=dfr)
test1_sum= r.summary(test1)
test2_sum= r.summary(test2)
print test1_sum
print test2_sum
这是针对nlme的代码，针对lme4的代码是：
test1=lme4.lmer(r.formula('Pupil~CoI*Time+(1|ID)'),data=dfr)
test2=lme4.lmer(r.formula('Pupil~CoI*measurement+(1|ID)'),data=dfr)
test1_sum= r.summary(test1)
test2_sum= r.summary(test2)
print test1_sum
print test2_sum
如果你想要一个包含数据和明确导入的代码片段，请参考这个IPython笔记本。
在所有情况下，我得到的输出信息非常多，其中有一段特别长的内容，看起来像是：
Data: structure(list(CoI = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L ......
我希望能得到一个更简洁的摘要，类似于：
Random effects:
Formula: ~1 | ID
(Intercept) Residual
StdDev: 0.2201214 0.1199874
Fixed effects: Pupil ~ CoI * measurement
Value Std.Error DF t-value p-value
(Intercept) 1.2068660 0.06369911 5769 18.946357 0
CoIhard -0.0394413 0.00629117 5769 -6.269306 0
measurement -0.0002743 0.00003207 5769 -8.554287 0
CoIhard:measurement 0.0005227 0.00004536 5769 11.524511 0
Correlation:
(Intr) CoIhrd msrmnt
CoIhard -0.049
measurement -0.060 0.612
CoIhard:measurement 0.043 -0.865 -0.707
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-9.86773055 -0.37638950 0.02085029 0.43203795 4.97364143
Number of Observations: 5784
Number of Groups: 12
（这个摘要包含在我得到的输出中，但在上面的内容之后有成千上万条记录才出现）我该如何实现这个呢？
1 个回答
2
正确的方法是使用 .rx2() 这个方法，有很多种不同的用法：
In [43]:
print test2_sum.names
Unable to unlink tempfile c:\docume~1\x60t\locals~1\temp\tmpnhw4n4
[1] "methTitle" "objClass" "devcomp" "isLmer" "useScale"
[6] "logLik" "family" "link" "ngrps" "coefficients"
[11] "sigma" "vcov" "varcor" "AICtab" "call"
[16] "residuals"
In [44]:
print test2_sum.rx2('vcov') # to access R type print out
Unable to unlink tempfile c:\docume~1\x60t\locals~1\temp\tmpebn3f1
4 x 4 Matrix of class "dpoMatrix"
(Intercept) CoIhard measurement CoIhard:measurement
(Intercept) 93253.4275120 -80.6588422 -0.503069702 0.503069702
CoIhard -80.6588422 161.3176844 0.503069702 -1.006139404
measurement -0.5030697 0.5030697 0.004192248 -0.004192248
CoIhard:measurement 0.5030697 -1.0061394 -0.004192248 0.008384495
In [45]:
print test2_sum.rx2('varcor') # to access R type print out
Unable to unlink tempfile c:\docume~1\x60t\locals~1\temp\tmpcad6ld
Groups Name Std.Dev.
ID (Intercept) 1057.39
Residual 242.24
In [46]:
list(test2_sum.rx2('varcor')) # to get the values
Out[46]:
[<Matrix - Python:0x0782CEB8 / R:0x0E97FB28>
[1118073.223847]]
In [47]:
list(test2_sum.rx2('varcor')[0]) # to get the values
Out[47]:
[1118073.2238471208]
通过跳过 calls 和 residuals，你可以去掉大部分内容，试试这个：
for i, v in enumerate(list(test2_sum.names)):
if v not in ['call', 'residuals']:
print '%s========================================================='%i, v
print test2_sum.rx2(v)
补充说明：
我认为访问 lme4 结果中的 tTable（使用 rpy2）的最佳方法是将其转换为 pandas 的 DataFrame：
In [73]:
print com.convert_robj(test2_sum.rx2('tTable'))
Value Std.Error DF t-value p-value
(Intercept) 2480.515542 305.374210 5769 8.122872 5.521357e-16
CoIhard -90.840336 12.701090 5769 -7.152169 9.602962e-13
measurement -0.288709 0.064748 5769 -4.458998 8.390496e-06
CoIhard:measurement 1.049136 0.091567 5769 11.457595 4.546122e-30
[4 rows x 5 columns]
虽然 print 输出和 R 的打印结果不完全一样，但其实很容易就能做到：
In [87]:
print test2_sum.rx2('tTable').__str__().replace('\r\n\r\n', '\n')
Value Std.Error DF t-value p-value
(Intercept) 2480.5155423 305.37420990 5769 8.122872 5.521357e-16
CoIhard -90.8403359 12.70108989 5769 -7.152169 9.602962e-13
measurement -0.2887093 0.06474757 5769 -4.458998 8.390496e-06
CoIhard:measurement 1.0491363 0.09156689 5769 11.457595 4.546122e-30