问题:标准化大熊猫中的数据

假设我有一个熊猫数据框df

我想计算数据框的列均值。

这很简单:

df.apply(average) 

然后按列范围max(col)-min(col)。这又很容易:

df.apply(max) - df.apply(min)

现在,对于每个元素,我要减去其列的均值并除以其列的范围。我不确定该怎么做

非常感谢任何帮助/指针。

Suppose I have a pandas data frame df:

I want to calculate the column wise mean of a data frame.

This is easy:

df.apply(average) 

then the column wise range max(col) – min(col). This is easy again:

df.apply(max) - df.apply(min)

Now for each element I want to subtract its column’s mean and divide by its column’s range. I am not sure how to do that

Any help/pointers are much appreciated.


回答 0

In [92]: df
Out[92]:
           a         b          c         d
A  -0.488816  0.863769   4.325608 -4.721202
B -11.937097  2.993993 -12.916784 -1.086236
C  -5.569493  4.672679  -2.168464 -9.315900
D   8.892368  0.932785   4.535396  0.598124

In [93]: df_norm = (df - df.mean()) / (df.max() - df.min())

In [94]: df_norm
Out[94]:
          a         b         c         d
A  0.085789 -0.394348  0.337016 -0.109935
B -0.463830  0.164926 -0.650963  0.256714
C -0.158129  0.605652 -0.035090 -0.573389
D  0.536170 -0.376229  0.349037  0.426611

In [95]: df_norm.mean()
Out[95]:
a   -2.081668e-17
b    4.857226e-17
c    1.734723e-17
d   -1.040834e-17

In [96]: df_norm.max() - df_norm.min()
Out[96]:
a    1
b    1
c    1
d    1
In [92]: df
Out[92]:
           a         b          c         d
A  -0.488816  0.863769   4.325608 -4.721202
B -11.937097  2.993993 -12.916784 -1.086236
C  -5.569493  4.672679  -2.168464 -9.315900
D   8.892368  0.932785   4.535396  0.598124

In [93]: df_norm = (df - df.mean()) / (df.max() - df.min())

In [94]: df_norm
Out[94]:
          a         b         c         d
A  0.085789 -0.394348  0.337016 -0.109935
B -0.463830  0.164926 -0.650963  0.256714
C -0.158129  0.605652 -0.035090 -0.573389
D  0.536170 -0.376229  0.349037  0.426611

In [95]: df_norm.mean()
Out[95]:
a   -2.081668e-17
b    4.857226e-17
c    1.734723e-17
d   -1.040834e-17

In [96]: df_norm.max() - df_norm.min()
Out[96]:
a    1
b    1
c    1
d    1

回答 1

如果您不介意导入sklearn库,我建议您使用博客上介绍的方法。

import pandas as pd
from sklearn import preprocessing

data = {'score': [234,24,14,27,-74,46,73,-18,59,160]}
cols = data.columns
df = pd.DataFrame(data)
df

min_max_scaler = preprocessing.MinMaxScaler()
np_scaled = min_max_scaler.fit_transform(df)
df_normalized = pd.DataFrame(np_scaled, columns = cols)
df_normalized

If you don’t mind importing the sklearn library, I would recommend the method talked on this blog.

import pandas as pd
from sklearn import preprocessing

data = {'score': [234,24,14,27,-74,46,73,-18,59,160]}
cols = data.columns
df = pd.DataFrame(data)
df

min_max_scaler = preprocessing.MinMaxScaler()
np_scaled = min_max_scaler.fit_transform(df)
df_normalized = pd.DataFrame(np_scaled, columns = cols)
df_normalized

回答 2

您可以使用apply它,它有点整洁:

import numpy as np
import pandas as pd

np.random.seed(1)

df = pd.DataFrame(np.random.randn(4,4)* 4 + 3)

          0         1         2         3
0  9.497381  0.552974  0.887313 -1.291874
1  6.461631 -6.206155  9.979247 -0.044828
2  4.276156  2.002518  8.848432 -5.240563
3  1.710331  1.463783  7.535078 -1.399565

df.apply(lambda x: (x - np.mean(x)) / (np.max(x) - np.min(x)))

          0         1         2         3
0  0.515087  0.133967 -0.651699  0.135175
1  0.125241 -0.689446  0.348301  0.375188
2 -0.155414  0.310554  0.223925 -0.624812
3 -0.484913  0.244924  0.079473  0.114448

此外,groupby如果您选择相关列,它也可以与配合使用:

df['grp'] = ['A', 'A', 'B', 'B']

          0         1         2         3 grp
0  9.497381  0.552974  0.887313 -1.291874   A
1  6.461631 -6.206155  9.979247 -0.044828   A
2  4.276156  2.002518  8.848432 -5.240563   B
3  1.710331  1.463783  7.535078 -1.399565   B


df.groupby(['grp'])[[0,1,2,3]].apply(lambda x: (x - np.mean(x)) / (np.max(x) - np.min(x)))

     0    1    2    3
0  0.5  0.5 -0.5 -0.5
1 -0.5 -0.5  0.5  0.5
2  0.5  0.5  0.5 -0.5
3 -0.5 -0.5 -0.5  0.5

You can use apply for this, and it’s a bit neater:

import numpy as np
import pandas as pd

np.random.seed(1)

df = pd.DataFrame(np.random.randn(4,4)* 4 + 3)

          0         1         2         3
0  9.497381  0.552974  0.887313 -1.291874
1  6.461631 -6.206155  9.979247 -0.044828
2  4.276156  2.002518  8.848432 -5.240563
3  1.710331  1.463783  7.535078 -1.399565

df.apply(lambda x: (x - np.mean(x)) / (np.max(x) - np.min(x)))

          0         1         2         3
0  0.515087  0.133967 -0.651699  0.135175
1  0.125241 -0.689446  0.348301  0.375188
2 -0.155414  0.310554  0.223925 -0.624812
3 -0.484913  0.244924  0.079473  0.114448

Also, it works nicely with groupby, if you select the relevant columns:

df['grp'] = ['A', 'A', 'B', 'B']

          0         1         2         3 grp
0  9.497381  0.552974  0.887313 -1.291874   A
1  6.461631 -6.206155  9.979247 -0.044828   A
2  4.276156  2.002518  8.848432 -5.240563   B
3  1.710331  1.463783  7.535078 -1.399565   B


df.groupby(['grp'])[[0,1,2,3]].apply(lambda x: (x - np.mean(x)) / (np.max(x) - np.min(x)))

     0    1    2    3
0  0.5  0.5 -0.5 -0.5
1 -0.5 -0.5  0.5  0.5
2  0.5  0.5  0.5 -0.5
3 -0.5 -0.5 -0.5  0.5

回答 3

稍作修改自:Python Pandas数据框:归一化0.01和0.99之间的数据?但是从一些评论中认为这是相关的(抱歉,如果考虑重新发布…)

我想要自定义归一化,因为基准或z分数的常规百分位数不够。有时我知道总体的可行最大值和最小值是多少,因此除了我的样本或其他中点之外,还想对其进行定义!这通常对于重新缩放和规范化神经网络的数据很有用,因为您可能希望所有输入都在0到1之间,但是某些数据可能需要以更自定义的方式进行缩放…因为百分位数和标准差假设您的样本覆盖了人口,但有时我们知道这是不对的。在可视化热图中的数据时,这对我也非常有用。因此,我构建了一个自定义函数(在此处的代码中使用了额外的步骤,以使其更具可读性):

def NormData(s,low='min',center='mid',hi='max',insideout=False,shrinkfactor=0.):    
    if low=='min':
        low=min(s)
    elif low=='abs':
        low=max(abs(min(s)),abs(max(s)))*-1.#sign(min(s))
    if hi=='max':
        hi=max(s)
    elif hi=='abs':
        hi=max(abs(min(s)),abs(max(s)))*1.#sign(max(s))

    if center=='mid':
        center=(max(s)+min(s))/2
    elif center=='avg':
        center=mean(s)
    elif center=='median':
        center=median(s)

    s2=[x-center for x in s]
    hi=hi-center
    low=low-center
    center=0.

    r=[]

    for x in s2:
        if x<low:
            r.append(0.)
        elif x>hi:
            r.append(1.)
        else:
            if x>=center:
                r.append((x-center)/(hi-center)*0.5+0.5)
            else:
                r.append((x-low)/(center-low)*0.5+0.)

    if insideout==True:
        ir=[(1.-abs(z-0.5)*2.) for z in r]
        r=ir

    rr =[x-(x-0.5)*shrinkfactor for x in r]    
    return rr

这将采用熊猫系列,甚至只是一个列表,并将其标准化为您指定的低点,中点和高点。还有一个缩小因素!使您可以缩小端点0和1之外的数据的比例(在matplotlib中组合颜色图时,我必须这样做:使用Matplotlib单个pcolormesh中使用多个颜色图)样本中具有[-5,1,10]的值,但要基于-7到7(因此,大于7的任何值,我们的“ 10”有效地视为7)以2为中点进行归一化但将其缩小以适合256 RGB色彩图:

#In[1]
NormData([-5,2,10],low=-7,center=1,hi=7,shrinkfactor=2./256)
#Out[1]
[0.1279296875, 0.5826822916666667, 0.99609375]

它也可以将您的数据完全翻过来……这似乎很奇怪,但是我发现它对于热图很有用。假设您想使用深色来表示接近0的值,而不是高/低。您可以基于归一化数据的热图,其中Insideout = True:

#In[2]
NormData([-5,2,10],low=-7,center=1,hi=7,insideout=True,shrinkfactor=2./256)
#Out[2]
[0.251953125, 0.8307291666666666, 0.00390625]

因此,现在最接近中心的“ 2”(定义为“ 1”)是最大值。

无论如何,如果您希望以其他可能对您有用的应用程序重新缩放数据的方式,我认为我的应用程序很重要。

Slightly modified from: Python Pandas Dataframe: Normalize data between 0.01 and 0.99? but from some of the comments thought it was relevant (sorry if considered a repost though…)

I wanted customized normalization in that regular percentile of datum or z-score was not adequate. Sometimes I knew what the feasible max and min of the population were, and therefore wanted to define it other than my sample, or a different midpoint, or whatever! This can often be useful for rescaling and normalizing data for neural nets where you may want all inputs between 0 and 1, but some of your data may need to be scaled in a more customized way… because percentiles and stdevs assumes your sample covers the population, but sometimes we know this isn’t true. It was also very useful for me when visualizing data in heatmaps. So i built a custom function (used extra steps in the code here to make it as readable as possible):

def NormData(s,low='min',center='mid',hi='max',insideout=False,shrinkfactor=0.):    
    if low=='min':
        low=min(s)
    elif low=='abs':
        low=max(abs(min(s)),abs(max(s)))*-1.#sign(min(s))
    if hi=='max':
        hi=max(s)
    elif hi=='abs':
        hi=max(abs(min(s)),abs(max(s)))*1.#sign(max(s))

    if center=='mid':
        center=(max(s)+min(s))/2
    elif center=='avg':
        center=mean(s)
    elif center=='median':
        center=median(s)

    s2=[x-center for x in s]
    hi=hi-center
    low=low-center
    center=0.

    r=[]

    for x in s2:
        if x<low:
            r.append(0.)
        elif x>hi:
            r.append(1.)
        else:
            if x>=center:
                r.append((x-center)/(hi-center)*0.5+0.5)
            else:
                r.append((x-low)/(center-low)*0.5+0.)

    if insideout==True:
        ir=[(1.-abs(z-0.5)*2.) for z in r]
        r=ir

    rr =[x-(x-0.5)*shrinkfactor for x in r]    
    return rr

This will take in a pandas series, or even just a list and normalize it to your specified low, center, and high points. also there is a shrink factor! to allow you to scale down the data away from endpoints 0 and 1 (I had to do this when combining colormaps in matplotlib:Single pcolormesh with more than one colormap using Matplotlib) So you can likely see how the code works, but basically say you have values [-5,1,10] in a sample, but want to normalize based on a range of -7 to 7 (so anything above 7, our “10” is treated as a 7 effectively) with a midpoint of 2, but shrink it to fit a 256 RGB colormap:

#In[1]
NormData([-5,2,10],low=-7,center=1,hi=7,shrinkfactor=2./256)
#Out[1]
[0.1279296875, 0.5826822916666667, 0.99609375]

It can also turn your data inside out… this may seem odd, but I found it useful for heatmapping. Say you want a darker color for values closer to 0 rather than hi/low. You could heatmap based on normalized data where insideout=True:

#In[2]
NormData([-5,2,10],low=-7,center=1,hi=7,insideout=True,shrinkfactor=2./256)
#Out[2]
[0.251953125, 0.8307291666666666, 0.00390625]

So now “2” which is closest to the center, defined as “1” is the highest value.

Anyways, I thought my application was relevant if you’re looking to rescale data in other ways that could have useful applications to you.


回答 4

这是按列进行的方式:

[df[col].update((df[col] - df[col].min()) / (df[col].max() - df[col].min())) for col in df.columns]

This is how you do it column-wise:

[df[col].update((df[col] - df[col].min()) / (df[col].max() - df[col].min())) for col in df.columns]

声明:本站所有文章,如无特殊说明或标注,均为本站原创发布。任何个人或组织,在未征得本站同意时,禁止复制、盗用、采集、发布本站内容到任何网站、书籍等各类媒体平台。如若本站内容侵犯了原著者的合法权益,可联系我们进行处理。