如何每隔n个项目拆分一个列表

Question

我试图每隔 5 个项目拆分一个列表，然后删除接下来的两个项目（'nan'）。我曾尝试使用 List[:5]，但这似乎无法循环使用。所需的输出是：[['1','2','3','4','5'],['1','2','3','4','5'], ['1','2','3','4','5'],['1','2','3','4','5']]List = ['1','2','3','4','5','nan','nan','1','2','3','4','5','nan','nan','1','2','3','4','5','nan','nan','1','2','3','4','5','nan','nan']for i in List:    # split first 5 items    # delete next two items# Desired output:# [['1','2','3','4','5'],['1','2','3','4','5'],['1','2','3','4','5'],['1','2','3','4','5']]

Answer 1

有很多方法可以做到这一点。我建议步进 7 然后拼接 5。

data = ['1','2','3','4','5','nan','nan','1','2','3','4','5','nan','nan','1','2','3','4','5','nan','nan','1','2','3','4','5','nan','nan']

# Step by 7 and keep the first 5

chunks = [data[i:i+5] for i in range(0, len(data), 7)]

print(*chunks, sep='\n')

输出：

['1', '2', '3', '4', '5']

['1', '2', '3', '4', '5']

['1', '2', '3', '4', '5']

['1', '2', '3', '4', '5']

Answer 2

警告：确保列表遵循您所说的规则，每 5 项 2 nan 之后。

此循环会将前 5 个项目添加为列表，并删除前 7 个项目。

lst = ['1','2','3','4','5','nan','nan','1','2','3','4','5','nan','nan','1','2','3','4','5','nan','nan','1','2','3','4','5','nan','nan']

output = []

while True:

    if len(lst) <= 0:

        break

    output.append(lst[:5])

    del lst[:7]

print(output) # [['1', '2', '3', '4', '5'], ['1', '2', '3', '4', '5'], ['1', '2', '3', '4', '5'], ['1', '2', '3', '4', '5']]

Answer 3

我喜欢拼接的答案。

这是我的 2 美分。

# changed var name away from var type

myList = ['1','2','3','4','5','nan','nan','1','2','3','4','10','nan','nan','1','2','3','4','15','nan','nan','1','2','3','4','20','nan','nan']

newList = []   # declare new list of lists to create

addItem = []   # declare temp list

myIndex = 0    # declare temp counting variable

for i in myList:

    myIndex +=1

    if myIndex==6:

        nothing = 0  #do nothing

    elif myIndex==7: #add sub list to new list and reset variables

        if len(addItem)>0:

            newList.append(list(addItem))

            addItem=[]

            myIndex = 0

    else:

       addItem.append(i)

#output

print(newList)

Answer 4

如果列表不遵循您提到的规则但您希望始终在 NAN 之间拆分序列，则直接解决方案：

result, temp = [], []

for item in lst:

    if item != 'nan':

        temp.append(item)

    elif temp:

        result.append(list(temp))

        temp = []