使用np.select/ValueError:-1不在范围内，获取第一个小于x的值

Question

我在代码的np.select部分遇到了一个问题，并复制了一个最小可复制示例，以寻求一些建议，说明为什么返回ValueError: -1 is not in range而不是nan

import numpy as np
import pandas as pd
test = {'number' : [1,2,3,4,5,6]}
df = pd.DataFrame(data=test)
print(df)
number = 1
#check row index of first value less than 'number'
print((np.abs(df['number']-number)).values.argmin()-1)
conditions = [number <= df['number'][3], number > df['number'][3]]
selection = [df['number'][(np.abs(df['number']-number)).values.argmin()-1], 'ignore'] # get first value in df['number'] column less than 'number' variable
answer = np.select(conditions, selection, default=np.nan)
print(answer)

当number = 1时使用df['number'][3]，我希望返回nan，因为位于df['number'][3]中的值是4，并且尽管number = 1小于4，但在df['number']中的行索引上方没有值为1的行

相反，我得到的是ValueError: -1 is not in range而不是nan

Answer 1

修改代码以避免上述错误-这只是原始代码的典型版本，但有轻微修改和if statement:

import numpy as np
import pandas as pd
test = {'number' : [1,2,3,4,5,6]}
df = pd.DataFrame(data=test)
print(df)
number = 1
#check row index of first value less than 'number'
row_index_less =(np.abs(df['number']-number)).values.argmin()-1
print(row_index_less)
if row_index_less>-1:
conditions = [number <= df['number'][3], number > df['number'][3]]
selection = [df['number'][row_index_less], 'ignore'] 
# get first value in df['number'] column less than 'number' variable
answer = np.select(conditions, selection, default=np.nan)
else
answer = np.nan
print(answer)