I would like to know if it is possible to get more numbers after the comma with classification_report from sklearn (scikit).
atm it looks like this:
precision recall f1-score support
1 0.61 0.73 0.67 71194
2 0.64 0.33 0.43 13877
3 0.56 0.59 0.57 61591
4 0.64 0.51 0.57 13187
5 0.66 0.69 0.67 57530
6 0.54 0.06 0.11 2391
7 0.54 0.40 0.46 30223
avg / total 0.60 0.60 0.60 249993
I don't think it is possible with that method, but maybe someone had the same idea (probably).
I know that sklearn.metrics.precision_score exists, though the classification_report is such a nice way to display all the results at once.
Not possible according to the source code. See lines 819 and 830, format strings are hardcoded to %0.2f. If you really want it, just change it in your local file sklearn/metrics/metrics.py. Better yet, add an argument to classification_report with a precision number and use that. And submit your patch to the project!
Related
I am working on a churn classification with 3 classes 0, 1,2 but want to optimize class 0 and 1 for recall, does that mean sklearn needs to take classes 0 & 1 to be the positive classes. How can I explicitly mention for which class do I want to optimise recall , if that is not possible should I consider renaming the classes in an ascending order so that 1, 2 are default positive?
precision recall f1-score support
0 0.71 0.18 0.28 2611
1 0.57 0.54 0.56 5872
2 0.70 0.88 0.78 8913
accuracy 0.66 17396
macro avg 0.66 0.53 0.54 17396
weighted avg 0.66 0.66 0.63 17396
Here is the code I am using for reference (although I need more of an understanding of how to optimize for recall for only 0, 1 class here)
param_test1={'learning_rate':(0.05,0.1),'max_depth':(3,5)}
estimator=GridSearchCV(estimator=GradientBoostingClassifier(loss='deviance',subsample=0.8,random_state=10,
n_estimators=200),param_grid=param_test1,cv=2, refit='recall_score')
estimator.fit(df[predictors],df[target])
I want to pass the parameters in my def to inputs in pandas loc but I am not sure how to do so, as loc requires defined labels as inputs. Or is there any other way I can perform Excel INDEX MATCH equivalent in Python but not using loc? Many thanks!
Below please find my code:
def get_correl_diff_tenor(p1, p2):
correl = IRCorrMatrix.loc['p1', 'p2']
return correl
p1 and p2 in loc['p1', 'p2'] refer to the tenor pairs for calling the corresponding correlation value in the matrix below.
IRCorrMatrix is shown below, which is a correlation matrix defined by tenor pairs.
2w 1m 3m 6m 1y
Tenor
2w 1.00 0.73 0.64 0.57 0.44
1m 0.73 1.00 0.78 0.67 0.50
3m 0.64 0.78 1.00 0.85 0.66
6m 0.57 0.67 0.85 1.00 0.81
1y 0.44 0.50 0.66 0.81 1.00
IIUC remove '' from 'p1', 'p2' for pass variables from function:
IRCorrMatrix.loc[p1, p2]
I called describe on one column of a dataframe and ended up with the following output,
count 1.048575e+06
mean 8.232821e+01
std 2.859016e+02
min 0.000000e+00
25% 3.000000e+00
50% 1.400000e+01
75% 6.000000e+01
max 8.599700e+04
What parameter do I pass to get meaningful integer values. What I mean is when I check the SQL count its about 43 million. All the other values are also different.Can someone help me understand what this conversion means and how do I get float rounded to 2 decimal places. I'm new to Pandas.
You can directly use round() and pass the number of decimals you want as argument
# importing pandas as pd
import pandas as pd
# importing numpy as np
import numpy as np
# setting the seed to create the dataframe
np.random.seed(25)
# Creating a 5 * 4 dataframe
df = pd.DataFrame(np.random.random([5, 4]), columns =["A", "B", "C", "D"])
# rounding describe
df.describe().round(2)
A B C D
count 5.00 5.00 5.00 5.00
mean 0.52 0.47 0.38 0.42
std 0.21 0.23 0.19 0.29
min 0.33 0.12 0.16 0.11
25% 0.41 0.37 0.28 0.19
50% 0.45 0.58 0.37 0.44
75% 0.56 0.59 0.40 0.52
max 0.87 0.70 0.68 0.84
DOCS
There are two ways to control the output of pandas, either by controlling it or by using apply.
pd.set_option('display.float_format', lambda x: '%.5f' % x)
df['X'].describe().apply("{0:.5f}".format)
I'm using classification_report function which is supposed to return a dictionary, but it doesn't return a well-structured one. So how can I print it well in jupyter notebook?
from sklearn.metrics import classification_report
classification_report(test_preds, y_test)
Here is my output:
' precision recall f1-score support\n\n 0 0.81 0.97 0.88 1653\n
1 0.76 0.28 0.41 533\n\n accuracy 0.80 2186\n macro avg
0.78 0.63 0.65 2186\nweighted avg 0.79 0.80 0.77 2186\n'
use print(classification_report(test_preds, y_test))
I am working with two numeric data.frames, both with 13803obs and 13803 variables. Their col- and rownames are identical however their entries are different. What I want to do is create a new data.frame where I have subtracted df2 values with df1 values.
"Formula" would be this, df1(entri-values) - df2(entri-values) = df3 difference. In other words, the purpose is to find the difference between all entries.
My problem illustrated here.
DF1
[GENE128] [GENE271] [GENE2983]
[GENE231] 0.71 0.98 0.32
[GENE128] 0.23 0.61 0.90
[GENE271] 0.87 0.95 0.63
DF2
[GENE128] [GENE271] [GENE2983]
[GENE231] 0.70 0.94 0.30
[GENE128] 0.25 0.51 0.80
[GENE271] 0.82 0.92 0.60
NEW DF3
[GENE128] [GENE271] [GENE2983]
[GENE231] 0.01 0.04 0.02
[GENE128] -.02 0.10 0.10
[GENE271] 0.05 0.03 0.03
So, in DF3 the values are the difference between DF1 and DF2 for each entry.
DF1(GENE231) - DF2(GENE231) = DF3(DIFFERENCE-GENE231)
DF1(GENE271) - DF2(GENE271) = DF3(DIFFERENCE-GENE271)
and so on...
Help would be much appreciated!
Kind regards,
Harebell