python 在列表的列表中循环 x + 1 次直到编号 y

python loop x + 1 times in a list of list until number y

我想要的是 num 中的列表每次循环 x + 1 直到生成 y(并停止循环),这是一个很大的数字。

def loop_num(y):
    num = []
    num.append([1])
    num.append([2,3])
    num.append([4,5,6]) 
    num.append([7,8,9,10]) 
    ... #stop append when y in the appended list
    #if y = 9, then `append [7,8]` and `return num`
    return num


# something like[[1 item], [2items], [3items], ...]
# the number append to the list can be a random number or ascending integers. 

抱歉不清楚

我不确定你的问题是什么。但我假设你已经做过类似的事情。

x=[[1], [2,3], [4,5,6]]
b=[str(a)+' items' for a in [j for i in x for j in i]]

而你要找的就是这个。

c=max([j for i in x for j in i])

做这个。

z=[]
z.append(str(c)+' items')

两个 itertools.count 对象应该做你想做的事:

from itertools import count

def loop_num(y):
    counter, x = count(1), count(1)
    n = 0
    while n < y:
        num = []
        for i in range(next(x)):
            num.append(next(counter))
            if num[-1] == y:
                break
        yield num
        n = num[-1]

输出:

>>> list(loop_num(100))
[[1],
 [2, 3],
 [4, 5, 6],
 [7, 8, 9, 10],
 [11, 12, 13, 14, 15],
 [16, 17, 18, 19, 20, 21],
 [22, 23, 24, 25, 26, 27, 28],
 [29, 30, 31, 32, 33, 34, 35, 36],
 [37, 38, 39, 40, 41, 42, 43, 44, 45],
 [46, 47, 48, 49, 50, 51, 52, 53, 54, 55],
 [56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66],
 [67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78],
 [79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91],
 [92, 93, 94, 95, 96, 97, 98, 99, 100]]
def loop_num(x):
    i=1
    cnt=0
    sum=0
    while sum<x:

        sum+=i
        cnt=cnt+1
        i=i+1

    num=[ [] for x in range(cnt)]

    count=0

    sz=1
    init=1
    while(count<cnt):
        cur=1
        while(cur<=sz):
            num[count].append(init)
            init=init+1
            cur=cur+1
        count=count+1
        sz=sz+1;
    return num

从一个 python 文件你可以从命令行 运行 它(比如文件名是 test.py)

python -c 'import test; print test.loop_num(55)'