如何從給定數量的輸入或Python中的列表創建具有固定長度的子列表？

Question

我想從Python中給定的輸入數量創建具有固定列表長度的子列表。

例如，我的輸入是： ['a','b','c',......'z'] ...然后，我想將這些值放在幾個列表中。 每個列表的長度應為6。所以我想要這樣的東西：

first list = ['a','b','c','d','e','f']

second list = ['g','h','i','j','k','l']

last list = [' ',' ',' ',' ',' ','z' ]

我該如何實現？

Answer 1

這會將您的列表分為兩個等長（6）的列表：

>>> my_list = [1, 'ab', '', 'No', '', 'NULL', 2, 'bc', '','Yes' ,'' ,'Null']
>>> x = my_list[:len(my_list)//2]
>>> y = my_list[len(my_list)//2:]
>>> x
[1, 'ab', '', 'No', '', 'NULL']
>>> y
[2, 'bc', '', 'Yes', '', 'Null']

如果要將列表拆分為許多較小的列表，請使用：

chunks = [my_list[x:x+size] for x in range(0, len(my_list), size)]

其中size是所需的較小列表的大小，例如：

>>> size = 2
>>> chunks = [my_list[x:x+size] for x in range(0, len(my_list), size)]
[[1, 'ab'], ['', 'No'], ['', 'NULL'], [2, 'bc'], ['', 'Yes'], ['', 'Null']]
>>> for item in chunks:
        print (item)
[1, 'ab']
['', 'No']
['', 'NULL']
[2, 'bc']
['', 'Yes']
['', 'Null']

Answer 2

這將返回一個二維列表“ b”，其中每個列表包含的塊數與大塊一樣多。

a = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
b = []
chunksize = 6
def get_list(a, chunk):
    return a[chunk*chunksize:chunk*chunksize+chunksize]
for i in range(int(len(a) / chunksize)):
    b.append(get_list(a,i))
print(b)

輸出：

[['a', 'b', 'c', 'd', 'e', 'f'], ['g', 'h', 'i', 'j', 'k', 'l'], ['m', 'n', 'o', 'p', 'q', 'r'], ['s', 't', 'u', 'v', 'w', 'x']]

Answer 3

您的輸入是一個字符串，您需要先用逗號將其分割，然后再將其進一步分割：

input_string = "1, 'ab', '', 'No', '', 'NULL', 2, 'bc', '','Yes' ,'' ,'Null'"
bits = input_string.split(',')
x,y = bits[:6],bits[6:] # divide by 6
x,y = bits[:len(bits)//2],bits[len(bits)//2:] # divide in half

Answer 4

最小的解決方案：

    x = ["a","b","c","d","e","f","g","h","i","j"]
    size = 3 (user input)
    for counter in range(0,len(x),size):
        print(x[counter:counter+size])