我有一个熊猫数据框,我希望将其分为3组。我知道使用train_test_split从sklearn.cross_validation
,一个可以在两个集(训练集和测试)分割数据。但是,我找不到将数据分为三组的任何解决方案。最好是,我想拥有原始数据的索引。
我知道一种解决方法是使用train_test_split
两次并以某种方式调整索引。但是,是否存在更标准/内置的方式将数据分为3组而不是2组?
我有一个熊猫数据框,我希望将其分为3组。我知道使用train_test_split从sklearn.cross_validation
,一个可以在两个集(训练集和测试)分割数据。但是,我找不到将数据分为三组的任何解决方案。最好是,我想拥有原始数据的索引。
我知道一种解决方法是使用train_test_split
两次并以某种方式调整索引。但是,是否存在更标准/内置的方式将数据分为3组而不是2组?
Answers:
脾气暴躁的解决方案。我们将首先对整个数据集进行洗牌(df.sample(frac = 1)),然后将数据集分为以下几部分:
In [305]: train, validate, test = np.split(df.sample(frac=1), [int(.6*len(df)), int(.8*len(df))])
In [306]: train
Out[306]:
A B C D E
0 0.046919 0.792216 0.206294 0.440346 0.038960
2 0.301010 0.625697 0.604724 0.936968 0.870064
1 0.642237 0.690403 0.813658 0.525379 0.396053
9 0.488484 0.389640 0.599637 0.122919 0.106505
8 0.842717 0.793315 0.554084 0.100361 0.367465
7 0.185214 0.603661 0.217677 0.281780 0.938540
In [307]: validate
Out[307]:
A B C D E
5 0.806176 0.008896 0.362878 0.058903 0.026328
6 0.145777 0.485765 0.589272 0.806329 0.703479
In [308]: test
Out[308]:
A B C D E
4 0.521640 0.332210 0.370177 0.859169 0.401087
3 0.333348 0.964011 0.083498 0.670386 0.169619
[int(.6*len(df)), int(.8*len(df))]
-是numpy.split()的indices_or_sections
数组。
这是一个小np.split()
用法演示-让我们将20个元素的数组拆分为以下部分:80%,10%,10%:
In [45]: a = np.arange(1, 21)
In [46]: a
Out[46]: array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20])
In [47]: np.split(a, [int(.8 * len(a)), int(.9 * len(a))])
Out[47]:
[array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]),
array([17, 18]),
array([19, 20])]
frac=1
指示sample()
函数返回所有(100%
或小数= 1.0
)行
np.random.seed(any_number)
在分割线前使用以获得每次运行相同的结果。其次,要使train:test:val::50:40:10
使用比例不相等[int(.5*len(dfn)), int(.9*len(dfn))]
。在此,第一个元素表示train
(0.5%)的尺寸,第二个元素表示val
(1-0.9 = 0.1%)的尺寸,两者之差表示test
(0.9-0.5 = 0.4%)的尺寸。如果我错了,请纠正我:)
编写函数来处理随机集创建的种子。您不应该依赖不会随机化集合的集合拆分。
import numpy as np
import pandas as pd
def train_validate_test_split(df, train_percent=.6, validate_percent=.2, seed=None):
np.random.seed(seed)
perm = np.random.permutation(df.index)
m = len(df.index)
train_end = int(train_percent * m)
validate_end = int(validate_percent * m) + train_end
train = df.iloc[perm[:train_end]]
validate = df.iloc[perm[train_end:validate_end]]
test = df.iloc[perm[validate_end:]]
return train, validate, test
np.random.seed([3,1415])
df = pd.DataFrame(np.random.rand(10, 5), columns=list('ABCDE'))
df
train, validate, test = train_validate_test_split(df)
train
validate
test
然而,一种方法将所述数据集成train
,test
,cv
与0.6
,0.2
,0.2
是使用该train_test_split
方法的两倍。
from sklearn.model_selection import train_test_split
x, x_test, y, y_test = train_test_split(xtrain,labels,test_size=0.2,train_size=0.8)
x_train, x_cv, y_train, y_cv = train_test_split(x,y,test_size = 0.25,train_size =0.75)
np.split()
。此外,它不需要额外的依赖sklearn
。
这是一个Python函数,可通过分层采样将Pandas数据帧分为训练,验证和测试数据帧。它通过train_test_split()
两次调用scikit-learn的函数来执行此拆分。
import pandas as pd
from sklearn.model_selection import train_test_split
def split_stratified_into_train_val_test(df_input, stratify_colname='y',
frac_train=0.6, frac_val=0.15, frac_test=0.25,
random_state=None):
'''
Splits a Pandas dataframe into three subsets (train, val, and test)
following fractional ratios provided by the user, where each subset is
stratified by the values in a specific column (that is, each subset has
the same relative frequency of the values in the column). It performs this
splitting by running train_test_split() twice.
Parameters
----------
df_input : Pandas dataframe
Input dataframe to be split.
stratify_colname : str
The name of the column that will be used for stratification. Usually
this column would be for the label.
frac_train : float
frac_val : float
frac_test : float
The ratios with which the dataframe will be split into train, val, and
test data. The values should be expressed as float fractions and should
sum to 1.0.
random_state : int, None, or RandomStateInstance
Value to be passed to train_test_split().
Returns
-------
df_train, df_val, df_test :
Dataframes containing the three splits.
'''
if frac_train + frac_val + frac_test != 1.0:
raise ValueError('fractions %f, %f, %f do not add up to 1.0' % \
(frac_train, frac_val, frac_test))
if stratify_colname not in df_input.columns:
raise ValueError('%s is not a column in the dataframe' % (stratify_colname))
X = df_input # Contains all columns.
y = df_input[[stratify_colname]] # Dataframe of just the column on which to stratify.
# Split original dataframe into train and temp dataframes.
df_train, df_temp, y_train, y_temp = train_test_split(X,
y,
stratify=y,
test_size=(1.0 - frac_train),
random_state=random_state)
# Split the temp dataframe into val and test dataframes.
relative_frac_test = frac_test / (frac_val + frac_test)
df_val, df_test, y_val, y_test = train_test_split(df_temp,
y_temp,
stratify=y_temp,
test_size=relative_frac_test,
random_state=random_state)
assert len(df_input) == len(df_train) + len(df_val) + len(df_test)
return df_train, df_val, df_test
以下是一个完整的工作示例。
考虑具有标签的数据集,您要在该标签上进行分层。该标签在原始数据集中具有自己的分布,例如75%foo
,15%bar
和10%baz
。现在,让我们使用60/20/20的比率将数据集分为训练,验证和测试子集,其中每个分割都保留标签的相同分布。请参见下图:
这是示例数据集:
df = pd.DataFrame( { 'A': list(range(0, 100)),
'B': list(range(100, 0, -1)),
'label': ['foo'] * 75 + ['bar'] * 15 + ['baz'] * 10 } )
df.head()
# A B label
# 0 0 100 foo
# 1 1 99 foo
# 2 2 98 foo
# 3 3 97 foo
# 4 4 96 foo
df.shape
# (100, 3)
df.label.value_counts()
# foo 75
# bar 15
# baz 10
# Name: label, dtype: int64
现在,让我们split_stratified_into_train_val_test()
从上方调用该函数以按照60/20/20的比例获取训练,验证和测试数据帧。
df_train, df_val, df_test = \
split_stratified_into_train_val_test(df, stratify_colname='label', frac_train=0.60, frac_val=0.20, frac_test=0.20)
三个数据框df_train
,df_val
和df_test
包含所有原始行,但它们的大小将遵循上述比率。
df_train.shape
#(60, 3)
df_val.shape
#(20, 3)
df_test.shape
#(20, 3)
此外,三个拆分中的每个拆分将具有相同的标签分布,即75%foo
,15%bar
和10%baz
。
df_train.label.value_counts()
# foo 45
# bar 9
# baz 6
# Name: label, dtype: int64
df_val.label.value_counts()
# foo 15
# bar 3
# baz 2
# Name: label, dtype: int64
df_test.label.value_counts()
# foo 15
# bar 3
# baz 2
# Name: label, dtype: int64
train_test_split
在划分为几组并且不编写一些其他代码之后不执行重新索引的情况下,使用起来非常方便。上面的最佳答案没有提到通过使用train_test_split
不改变分区大小来分隔两次不会给出最初想要的分区:
x_train, x_remain = train_test_split(x, test_size=(val_size + test_size))
然后,验证和测试集中x_remain中的部分发生变化,可以算作
new_test_size = np.around(test_size / (val_size + test_size), 2)
# To preserve (new_test_size + new_val_size) = 1.0
new_val_size = 1.0 - new_test_size
x_val, x_test = train_test_split(x_remain, test_size=new_test_size)
在这种情况下,将保存所有初始分区。