我有以下数据：
datetime price
2017-10-02 08:03:00 12877
2017-10-02 08:04:00 12877.5
2017-10-02 08:05:00 12879
2017-10-02 08:06:00 12875.5
2017-10-02 08:07:00 12875.5
2017-10-02 08:08:00 12878
2017-10-02 08:09:00 12878
2017-10-02 08:10:00 12878
2017-10-02 08:11:00 12881
2017-10-02 08:12:00 12882.5
2017-10-02 08:13:00 12884.5
2017-10-02 08:14:00 12882
2017-10-02 08:15:00 12880.5
2017-10-02 08:16:00 12881.5
2017-10-02 08:17:00 12879
2017-10-02 08:18:00 12879
2017-10-02 08:19:00 12880
2017-10-02 08:20:00 12878.5
我想找到分钟。 'datetime'的范围价格（范围由windows_size定义，可以是1/2/3等）使用：
df['MinPrice'] = df.ix[window_size:,'price']
它给出了窗口最后一行或使用
的价格df['MinPrice'] = df.ix[window_size:,'price'].min()
给出了所有列的最小值。
请建议如何获得分钟。窗口声明的特定行的值。
编辑： 预期结果如下：如果窗口大小为3，我想获得最小值。值为3行。所以在08:05:00我会得到12877而在08:06:00我会得到12875.5
答案 0 :(得分：1)
import java.util.*;
class Config{
int[] tabela;
Config(){
int[] blanks ={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,0};
tabela = blanks;
}
public Config(int arraydado[]){
tabela = arraydado;}
public void printTabela(){
for(int i = 0; i<16 ;i++)
System.out.print(tabela[i] + " ");
System.out.println();
}
}
public static Config move_left_new(Config dada){
int i;
int temp;
Config resultante = new Config(dada.tabela);
for(i = 0; i<16; i++){
if(resultante.tabela[i] == 0)
break;
}
if( i!=0 && i!= 4 && i!= 8 && i!=12){
temp = resultante.tabela[i-1];
resultante.tabela[i-1] = 0;
resultante.tabela[i] = temp;
}
return resultante;
}
public static void main(String[] args){
Scanner input = new Scanner(System.in);
int arr[] = new int[16];
for(int i=0; i<16; i++)
arr[i] = input.nextInt();
Config randomconfig = new Config(arr);
randomconfig.printTabela(); //original
Config changed = move_left_new(randomconfig);
randomconfig.printTabela(); //should be the same as before but isnt
changed.printTabela(); // moved as it should
}
将给出预期的结果：
df.rolling(window=3).apply(min).dropna()
答案 1 :(得分：1)
由于您看起来间隔为1分钟，因此您可能希望利用resample，这样您就可以使用日期时间来定义窗口
df.resample('3T',on='datetime').min()
datetime price
datetime
2017-10-02 08:03:00 2017-10-02 08:03:00 12877.0
2017-10-02 08:06:00 2017-10-02 08:06:00 12875.5
2017-10-02 08:09:00 2017-10-02 08:09:00 12878.0
2017-10-02 08:12:00 2017-10-02 08:12:00 12882.0
2017-10-02 08:15:00 2017-10-02 08:15:00 12879.0
2017-10-02 08:18:00 2017-10-02 08:18:00 12878.5
要将值设置回初始数据帧，请使用transform
df['minPrice'] = df.resample('3T',on='datetime').transform('min')
datetime price minPrice
0 2017-10-02 08:03:00 12877.0 12877.0
1 2017-10-02 08:04:00 12877.5 12877.0
2 2017-10-02 08:05:00 12879.0 12877.0
3 2017-10-02 08:06:00 12875.5 12875.5
4 2017-10-02 08:07:00 12875.5 12875.5
5 2017-10-02 08:08:00 12878.0 12875.5
6 2017-10-02 08:09:00 12878.0 12878.0
7 2017-10-02 08:10:00 12878.0 12878.0
8 2017-10-02 08:11:00 12881.0 12878.0
9 2017-10-02 08:12:00 12882.5 12882.0
10 2017-10-02 08:13:00 12884.5 12882.0
11 2017-10-02 08:14:00 12882.0 12882.0
12 2017-10-02 08:15:00 12880.5 12879.0
13 2017-10-02 08:16:00 12881.5 12879.0
14 2017-10-02 08:17:00 12879.0 12879.0
15 2017-10-02 08:18:00 12879.0 12878.5
16 2017-10-02 08:19:00 12880.0 12878.5
17 2017-10-02 08:20:00 12878.5 12878.5
答案 2 :(得分：0)
您可能希望保持数据框的长度相同：
df['Price_Low3'] = np.where(pd.isna(df.price.shift(periods=2)),df.price,df.price.rolling(3).min())
结果是：
datetime price Price_Low3
0 02/10/2017 08:03 12877.0 12877.0
1 02/10/2017 08:04 12877.5 12877.5
2 02/10/2017 08:05 12879.0 12877.0
3 02/10/2017 08:06 12875.5 12875.5
4 02/10/2017 08:07 12875.5 12875.5
5 02/10/2017 08:08 12878.0 12875.5
6 02/10/2017 08:09 12878.0 12875.5
7 02/10/2017 08:10 12878.0 12878.0
8 02/10/2017 08:11 12881.0 12878.0
9 02/10/2017 08:12 12882.5 12878.0
10 02/10/2017 08:13 12884.5 12881.0
11 02/10/2017 08:14 12882.0 12882.0
12 02/10/2017 08:15 12880.5 12880.5
13 02/10/2017 08:16 12881.5 12880.5
14 02/10/2017 08:17 12879.0 12879.0
15 02/10/2017 08:18 12879.0 12879.0
16 02/10/2017 08:19 12880.0 12879.0
17 02/10/2017 08:20 12878.5 12878.5