Retrieving the value of the row where the "time" value is 1 less for each row

Question

For each row, I need to find the row which is before 1 period(time column) and retrieve the value. To solve this, For loop is too slow... Is there any way to apply vector calculation for this? Or is there another way to speed up?

#making df
import pandas as pd
df = pd.DataFrame({'ID':[100001, 100002, 100003, 100004, 100005],
                'name':['Kim', 'Lee', 'Jeong', 'Park', 'Han'],
                'class':['H', 'W', 'S', "P", 'D'],
                'time':[1,2, 3, 4, 5],
               'value':[1111, 2222, 3333, 4444, 5555]
              })

Retrieving the value of the row where the "time" value is 1 less for each row : using for iteration

for i in range(0, len(df)):
    ID_row = df.at[i, 'ID']
    row_time = df.loc[df['ID']==ID_row]['time'].values[0]
    try:
        df.loc[df['ID']==ID_row, 'before_1'] = df.loc[df['time']==(row_time-1)]['value'].values[0]
    except:
        df.loc[df['ID']==ID_row, 'before_1'] = 0

Thank you!