45

如何在 Python 中生成数字序列“1,2,5,6,9,10......”等直到 100？我什至需要包含逗号（'，'），但这不是主要问题。

序列：从 1..100 开始的每个数字，可被 4 整除，余数为 1 或 2。

4

9 回答 9

41

1,2,5,6,9,10... 中的每个数字都可以被 4 整除，余数为 1 或 2。

>>> ','.join(str(i) for i in xrange(100) if i % 4 in (1,2))
'1,2,5,6,9,10,13,14,...'
于 2012-06-16T15:53:04.320 回答
20 
>>> ','.join('{},{}'.format(i, i + 1) for i in range(1, 100, 4))
'1,2,5,6,9,10,13,14,17,18,21,22,25,26,29,30,33,34,37,38,41,42,45,46,49,50,53,54,57,58,61,62,65,66,69,70,73,74,77,78,81,82,85,86,89,90,93,94,97,98'

That was a quick and quite dirty solution.

Now, for a solution that is suitable for different kinds of progression problems:

def deltas():
    while True:
        yield 1
        yield 3
def numbers(start, deltas, max):
    i = start
    while i <= max:
        yield i
        i += next(deltas)
print(','.join(str(i) for i in numbers(1, deltas(), 100)))

And here are similar ideas implemented using itertools:

from itertools import cycle, takewhile, accumulate, chain

def numbers(start, deltas, max):
    deltas = cycle(deltas)
    numbers = accumulate(chain([start], deltas))
    return takewhile(lambda x: x <= max, numbers)

print(','.join(str(x) for x in numbers(1, [1, 3], 100)))
于 2012-06-16T15:56:29.490 回答
13

Includes some guessing on the exact sequence you are expecting:

>>> l = list(range(1, 100, 4)) + list(range(2, 100, 4))
>>> l.sort()
>>> ','.join(map(str, l))
'1,2,5,6,9,10,13,14,17,18,21,22,25,26,29,30,33,34,37,38,41,42,45,46,49,50,53,54,57,58,61,62,65,66,69,70,73,74,77,78,81,82,85,86,89,90,93,94,97,98'

As one-liner:

>>> ','.join(map(str, sorted(list(range(1, 100, 4))) + list(range(2, 100, 4))))

(btw. this is Python 3 compatible)

于 2012-06-16T15:56:25.627 回答
2

这通过利用列表的 % 属性而不是增量来工作。

for num in range(1,100):
    if num % 4 == 1 or num % 4 ==2:
        n.append(num)
        continue
    pass
于 2016-01-03T14:42:43.020 回答
2

使用 numpy 和列表​​理解，你可以做到

import numpy as np
[num for num in np.arange(1,101) if (num%4 == 1 or num%4 == 2)]
于 2017-07-27T01:05:22.783 回答
0

假设您的序列在 1 和 3 之间交替递增

numbers = [1]
while numbers[-1] < 100:
    numbers.append(numbers[-1] + 1)
    numbers.append(numbers[-1] + 3)

print ', '.join(map(str, numbers))

如果您的序列不同，这可能更容易修改，但我认为 poke 或 BlaXpirit 比我的答案更好。

于 2012-06-16T16:05:27.470 回答
0

Assuming I've guessed the pattern correctly (alternating increments of 1 and 3), this should produce the desired result:

def sequence():
    res = []
    diff = 1
    x = 1
    while x <= 100:
        res.append(x)
        x += diff
        diff = 3 if diff == 1 else 1
    return ', '.join(res)
于 2012-06-16T15:57:55.190 回答
-1

编写一个函数，将数字作为参数并打印斐波那契数列直到该数字

def Series(n):  
    a = 0  
    b = 1  
    print(a)  
    print(b)  
    S = 0  
    for i in range(0,n):  
        if S <= n-1:  
            S = a + b  
            print(S)  
            a = b  
            b = S
于 2017-07-28T13:37:50.840 回答
-2

在 python 3.1 中，您可以通过某种方式生成列表

     lst=list(range(100))
     for i in range(100)
        print (lst[i],',',end='')

在 python 2.7 中，你可以这样做

     lst=range(100)
     for i in range(100)
        print lst[i]+','
于 2012-06-16T16:08:30.387 回答