Question

给定一个数据帧df

df = pandas.DataFrame(data=[1,0,0,1,1,1,0,1,0,1,1,1],columns = ['A'])

df

Out[20]: 
   A
0  1
1  0
2  0
3  1
4  1
5  1
6  0
7  1
8  0
9  1
10  1
11  1

我想找到间隔大于3的开始和结束索引。在这种情况下，我期望的是（3,5和9,11）

Answer 1

使用移位累积技巧来标记连续的组，然后使用groupby获取索引并根据您的条件进行过滤。

v = (df['A'] != df['A'].shift()).cumsum()
u = df.groupby(v)['A'].agg(['all', 'count'])
m = u['all'] & u['count'].ge(3)

df.groupby(v).apply(lambda x: (x.index[0], x.index[-1]))[m]

A
3     (3, 5)
7    (9, 11)
dtype: object

Answer 2

我没有明确认识Pandas，但我确实了解Python，因此将其视为一个小挑战：

def find_sub_in_list(my_list, sublist, greedy=True):
    matches = []
    results = []
    for item in range(len(my_list)):
        aux_list = my_list[item:]
        if len(sublist) > len(aux_list) or len(aux_list) == 0:
            break

        start_match = None
        end_pos = None

        if sublist[0] == my_list[item]:
            start_match = item
            for sub_item in range(len(sublist)):
                if sublist[sub_item] != my_list[item+sub_item]:
                    end_pos = False
        if end_pos == None and start_match != None:
            end_pos = start_match+len(sublist)
            matches.append([start_match, end_pos])

    if greedy:
        results = []
        for match in range(len(matches)-1):
            if matches[match][1] > matches[match+1][0]:
                results.append([matches[match][0], matches[match+1][1]])
            else:
                results.append(matches[match])

    else:
        results = matches

    return results


my_list = [1,1,1,0,1,1,0,1,1,1,1]

interval = 3
sublist = [1]*interval

matches = find_sub_in_list(my_list, sublist)
print(matches)

查找连续相等值的回合的开始结束索引

2 个答案: