递归函数很难向量化，因为每个输入都在时间t取决于在时间t-1的先前输入。
[问题更新如下，示例x_t=a x_{t-1}+b稍微复杂一些。]
.loc返回不同数据类型的问题
import pandas
df1 = pandas.DataFrame({'year':range(2020,2024),'a':range(3,7)})
# Set the initial value
t0 = min(df1.year)
df1.loc[df1.year==t0, "x"] = 0
当等式的右侧是pandas.core.series.series 时，此分配不起作用
for t in range (min(df1.year)+1, max(df1.year)+1):
df1.loc[df1.year==t, "x"] = df1.loc[df1.year==t-1,"x"] + df1.loc[df1.year==t-1,"a"]
print(df1)
# year a x
# 0 2020 3 0.0
# 1 2021 4 NaN
# 2 2022 5 NaN
# 3 2023 6 NaN
print(type(df1.loc[df1.year==t-1,"x"] + df1.loc[df1.year==t-1,"a"]))
# <class 'pandas.core.series.Series'>
当方程的右侧是一个numpy数组时，赋值有效
for t in range (min(df1.year)+1, max(df1.year)+1):
df1.loc[df1.year==t, "x"] = (df1.loc[df1.year==t-1,"x"] + df1.loc[df1.year==t-1,"a"]).unique()
#break
print(df1)
# year a x
# 0 2020 3 0.0
# 1 2021 4 3.0
# 2 2022 5 7.0
# 3 2023 6 12.0
print(type((df1.loc[df1.year==t-1,"x"] + df1.loc[df1.year==t-1,"a"]).unique()))
# <class 'numpy.ndarray'>
当.loc((选择使用年度索引时，赋值直接起作用
df2 = df.set_index("year").copy()
# Set the initial value
df2.loc[df2.index.min(), "x"] = 0
for t in range (df2.index.min()+1, df2.index.max()+1):
df2.loc[t, "x"] = df2.loc[t-1, "x"] + df2.loc[t-1,"a"]
#break
print(df2)
# a x
# year
# 2020 3 0.0
# 2021 4 3.0
# 2022 5 7.0
# 2023 6 12.0
print(type(df2.loc[t-1, "x"] + df2.loc[t-1,"a"]))
# <class 'numpy.float64'>
type(df1.loc[df1.year==t-1,"x"] + df1.loc[df1.year==t-1,"a"])是大熊猫而type(df2.loc[t-1, "x"] + df2.loc[t-1,"a"])是一个numpy浮点。为什么这些类型不同- 如果我不想在计算之前使用
set_index()。有没有比使用.unique()更好的方法来编写递归.loc()赋值
另请参阅：
- 递归上的相关问答任务
- [变异用户定义函数方法](https://pandas.pydata.org/pandas-docs/stable/user_guide/gotchas.html#mutating-使用用户定义的函数-udf方法(
使用乘法和加法分量的示例
我们的实际问题更为复杂，因为存在乘法和加法分量
import pandas
df3 = pandas.DataFrame({'year':range(2020,2024),'a':range(3,7), 'b':range(8,12)})
df3 = df3.set_index("year").copy()
# Set the initial value
df3.loc[df3.index.min(), "x"] = 0
for t in range (df3.index.min()+1, df3.index.max()+1):
df3.loc[t, "x"] = df3.loc[t-1, "x"] * df3.loc[t-1, "a"] + df3.loc[t-1, "b"]
#break
print(df3)
如果我不明白，很抱歉，你想要这个吗？
df1['x']= df1['a'].cumsum().shift().fillna(0)
print(df1)
输出：
year a x
0 2020 3 0.0
1 2021 4 3.0
2 2022 5 7.0
3 2023 6 12.0