Question

我试图从我的分析中不一致的大数据集中删除值。

以下是我开始使用的当前方法。

例如，假设我有一个由许多元素组成的数组a。

a = [30, 40, 200, 324, 8, 67, 789, 9, 567, 2143, 13]
idx = [(i,value) for i,value in enumerate(a) if value<=10]

print idx
>>> [(4, 8), (7, 9)]

我如何才能创建一个仅由其索引

组成的数组

print idx
>>> [4, 8]

Answer 1

不要在理解结果中包含该值：

idx = [i for i, v in enumerate(a) if v <= 10]

Answer 2

你几乎就在那里，只使用索引：

>>> a = [30, 40, 200, 324, 8, 67, 789, 9, 567, 2143, 13]
>>> idx = [i for i, value in enumerate(a) if value<=10]
>>> idx
[4, 7]

Answer 3

def Genom_Restriksiyon(sequence="MustafaUyar", splite="5", specifity=False):
        splite = int(splite)
        non_id_rstx_list = []
        id_rstx_list = []
        upper = 0
        _Treu_seq = ""
        for base in seq:
            if base != "\n":
                _Treu_seq += base
        while upper < splite:
            restriction = [_Treu_seq[i:i+splite] for i in range(upper, len(seq), splite)]
            position = 0
            for base in restriction:
                if specifity == False:
                    if len(base) == splite:
                        non_id_rstx_list.append(base)
                if specifity == True:
                    if len(base) == splite:
                        id_rstx_list.append((str(upper + (splite * position)) +
                                       " <"
                                       + base +
                                       "<= "
                                       + str(((splite * (position + 1)) - (splite - len(base))) + upper)))
                position += 1
            upper += 1
        if specifity == True:
            return id_rstx_list
        if specifity == False:
            return non_id_rstx_list

生成其元素为索引的数组

3 个答案: