总结从当前行到上一行的区别

Question

计算当前行与上一行的差，我有一个简单的数据集和下面的代码：

import pandas as pd

data = {'Month' : [1,2,3,4,5,6,7,8,9,10,11,12],
'Rainfall': [112,118,132,129,121,135,148,148,136,119,104,118]}

df = pd.DataFrame(data)

Rainfall = df["Rainfall"]

df['Changes'] = Rainfall.shift(-1) - Rainfall

df['Changes'] = df['Changes'].shift(1)

它显示更改（作为图片的左侧）。但是我只在乎变化是正，负还是零（如图片的右侧）

我试图添加一个IF条件，但这给了我错误：

if df['Changes'] > 0:
    df['Changes'] = df['Changes'].shift(1)

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

实现它的正确方法是什么？谢谢。

Answer 1

按字典将numpy.sign与map一起使用：

sudo -H -u user bash -c 'firefox -marionette'
sudo -H -u user bash -c 'firefox -marionette'
sudo -H -u user bash -c 'firefox -marionette'

使用numpy.select的另一种解决方案：

d = {1:'Positive', -1:'Negative',0:'Zero'}
df['Changes'] = np.sign(df['Changes'].shift(1)).map(d).fillna('')

print (df)
    Month  Rainfall   Changes
0       1       112        
1       2       118  Positive
2       3       132  Positive
3       4       129  Negative
4       5       121  Negative
5       6       135  Positive
6       7       148  Positive
7       8       148      Zero
8       9       136  Negative
9      10       119  Negative
10     11       104  Negative
11     12       118  Positive

编辑：

s = df['Changes'].shift(1)
df['Changes'] = np.select([s < 0, s > 0, s == 0],
                          ['Negative','Positive','Zero'], 
                           default='')

Answer 2

if条件不适用于pandas.Series。通过Changes创建了df['Changes'] = (Rainfall.shift(-1) - Rainfall) / Rainfall后，使用以下方法仅计算正数或负数-

df['Changes'] = np.where(df['Changes'] >0, 'positive', 'negative')
df['Changes'] = df['Changes'].shift(1)