Question

我有以下代码：

indices_to_remove= []
for i in range(0,len(df)):
        if (df.speed.values[i] <= 15 ):
            counter += 1
            if counter > 600:
               indices_to_remove.append(i)        

        else:
            counter= 0

df= df.drop (indices_to_remove, axis=0)

此代码的主要目标是遍历数据集中的所有行，并且如果有600多个连续行且速度值小于15，则该代码会将行索引添加到indexs_to_remove和那么所有这些行都将被删除。

Answer 1

您正在尝试并行执行两项操作，删除索引并计算小于15的600个连续值。我将把这两个想法分为两个步骤。

查找所有值小于15的索引
之后，计算连续的索引
如果其中有600多个索引，请执行删除操作

indices_to_remove= []

#Get all indexes to remove from the dataframe
for i in range(0,len(df)):
    if (df.speed.values[i] <= 15 ):
        indices_to_remove.append(i)

#Have a counter which keeps track of 600 consecutive indexes less than 15
counter = 0
max_counter = -1
for idx in range(len(indices_to_remove)-1):

    #If the indexes were consecutive, keep a counter
    if ((indices_to_remove[idx+1] - indices_to_remove[idx]) == 1):
        counter += 1

    #Else if non consecutive indexes are found, track the last maximum counter and reset the original counter
    else:
        if counter > max_counter:
            max_counter = counter
        counter = 0

if max_counter > 600:
    df = df.drop(indices_to_remove, axis=0)

Answer 2

这不是一个优雅的解决方案，而是基于您所拥有的功能：

indices_to_remove= []
indices_counter = []
for i in range(0,len(df)):

        if (df.speed.values[i] <= 15 ):
            counter += 1

            indices_counter.append(i)

            if counter > 600:
                commit = True

        elif commit:
            indices_to_remove.extend(indices_counter)
            indices_counter = []
            commit = False
            counter= 0

        else:
            indices_counter = []
            commit = False
            counter= 0

df= df.drop(indices_to_remove, axis=0)

Answer 3

建立一个索引字典，其中连续速度小于或等于15。

indices_dict = {}
k = 0
for i in range(0, len(df)):
    if (df.speed.values[i] <= 15 ):
        try:
            indices_dict[k].append(i)
        except KeyError:
            indices_dict[k] = [i]
    else:
        k += 1

lol_to_remove = [ v for k,v in indices_dict.items() if len(v)>= 600 ] # This is a list of lists (lol)
indices_to_remove = [i for v in lol_to_remove for i in v ] # flatten the list

df = df.drop(indices_to_remove, axis=0)

如何附加索引

3 个答案: