计算从当前行到上一行的差异,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)


它显示了更改(如左图所示)。但是我只在乎变化是正,负还是零(如图片的右侧)

python - 总结当前行与上一行的区别-LMLPHP

我试图添加一个IF条件,但它给了我错误:

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().


什么是实现它的正确方法?谢谢。

最佳答案

通过字典将numpy.signmap一起使用:

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


numpy.select的另一种解决方案:

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


编辑:

df['Changes'] = df['Changes'].shift(1)
bins = np.arange(-100, 100, step=5)
labels = ['{}-{}'.format(i, j) for i, j in zip(bins[:-1], bins[1:])]

df['Changes'] = pd.cut(df['Changes'], bins=bins, labels=labels)
print (df)
    Month  Rainfall Changes
0       1       112     NaN
1       2       118     0-5
2       3       132     0-5
3       4       129    -5-0
4       5       121    -5-0
5       6       135     0-5
6       7       148     0-5
7       8       148    -5-0
8       9       136    -5-0
9      10       119    -5-0
10     11       104    -5-0
11     12       118     0-5

关于python - 总结当前行与上一行的区别,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/54760821/

10-11 06:20