Python Pandas Dataframe：规范化0.01和0.99之间的数据吗？

问题描述

我正在尝试将数据框中的每个值限制在0.01到0.99之间

I am trying to bound every value in a dataframe between 0.01 and 0.99

我已成功使用以下方法对0到1之间的数据进行了规范化： .apply（lambda x：（x-x.min（））/（x.max（）-x.min（）））如下：

I have successfully normalised the data between 0 and 1 using: .apply(lambda x: (x - x.min()) / (x.max() - x.min())) as follows:

df = pd.DataFrame({'one' : ['AAL', 'AAL', 'AAPL', 'AAPL'], 'two' : [1, 1, 5, 5], 'three' : [4,4,2,2]})

df[['two', 'three']].apply(lambda x: (x - x.min()) / (x.max() - x.min()))

df

现在我想将所有值限制在0.01到0.99

Now I want to bound all values between 0.01 and 0.99

这是我尝试过的：

def bound_x(x):
    if x == 1:
        return x - 0.01
    elif x < 0.99:
        return x + 0.01

df[['two', 'three']].apply(bound_x)

df

但是我收到以下错误：

ValueError: ('The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().', u'occurred at index two')

推荐答案

有一个应用，错误，为此：

There's an app, err clip method, for that:

import pandas as pd
df = pd.DataFrame({'one' : ['AAL', 'AAL', 'AAPL', 'AAPL'], 'two' : [1, 1, 5, 5], 'three' : [4,4,2,2]})
df = df[['two', 'three']].apply(lambda x: (x - x.min()) / (x.max() - x.min()))
df = df.clip(lower=0.01, upper=0.99)

收益

    two  three
0  0.01   0.99
1  0.01   0.99
2  0.99   0.01
3  0.99   0.01

---

---

The problem with

df[['two', 'three']].apply(bound_x)

是 bound_x 通过了类似 df ['two']的系列，然后，如果x == 1 要求评估 x == 1 布尔上下文。 x == 1 是布尔级数，例如

is that bound_x gets passed a Series like df['two'] and then if x == 1 requires x == 1 be evaluated in a boolean context. x == 1 is a boolean Series like

In [44]: df['two'] == 1
Out[44]:
0    False
1    False
2     True
3     True
Name: two, dtype: bool

Python尝试将此Series简化为单个布尔值， True 或 False 。熊猫遵循。

Python tries to reduce this Series to a single boolean value, True or False. Pandas follows the NumPy convention of raising an error when you try to convert a Series (or array) to a bool.

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