@YTZ的回答很适合我。我必须生成一个从0到10000的列表，步长为0.01，由于舍入问题，在每次迭代中简单地添加0.01是行不通的。

因此，我采纳了@YTZ的建议，编写了如下函数:

import numpy as np


def generate_floating_numbers_in_range(start: int, end: int, step: float):
    """
    Generate a list of floating numbers within a specified range.

    :param start: range start
    :param end: range end
    :param step: range step
    :return:
    """
    numbers = np.linspace(start, end,(end-start)*int(1/step)+1).tolist()
    return [round(num, 2) for num in numbers]

你似乎在寻找range():

>>> x1=11
>>> x2=16
>>> range(x1, x2+1)
[11, 12, 13, 14, 15, 16]
>>> list1 = range(x1, x2+1)
>>> list1
[11, 12, 13, 14, 15, 16]

如果要增加0.5而不是1，那么:

>>> list2 = [x*0.5 for x in range(2*x1, 2*x2+1)]
>>> list2
[11.0, 11.5, 12.0, 12.5, 13.0, 13.5, 14.0, 14.5, 15.0, 15.5, 16.0]

在python中，你可以很容易地做到这一点

start=0
end=10
arr=list(range(start,end+1))
output: arr=[0,1,2,3,4,5,6,7,8,9,10]

或者你可以创建一个递归函数，返回一个数组到给定的数字:

ar=[]
def diff(start,end):
    if start==end:
        d.append(end)
        return ar
    else:
        ar.append(end)
        return diff(start-1,end) 

输出: 基于“增大化现实”技术=[10、9、8、7、6、5、4、3、2、1、0]

在python中使用列表理解式。既然你也想要16个。使用x2 + 1。Range函数不包括函数中的上限。

list=[x for x in range(x1, x2+1)]

虽然@Jared关于增加0.5步长的答案有效，但由于舍入问题，它无法用于其他步长:

np.arange(11, 17, 0.1).tolist()
# [11.0,11.1,11.2,11.299999999999999, ...   16.79999999999998, 16.899999999999977]

相反，我自己也需要这样的工作，不只是挣0.5美元:

# Example 11->16 step 0.5
s = 11
e = 16
step = 0.5
my_list = [round(num, 2) for num in np.linspace(s,e,(e-s)*int(1/step)+1).tolist()]
# [11.0, 11.5, 12.0, 12.5, 13.0, 13.5, 14.0, 14.5, 15.0, 15.5, 16.0]

# Example 0->1 step 0.1
s = 0
e = 1
step = 0.1
my_list = [round(num, 2) for num in np.linspace(s,e,(e-s)*int(1/step)+1).tolist()]
# [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]

如果你正在寻找适用于浮点类型的范围函数，那么这是一篇非常好的文章。

def frange(start, stop, step=1.0):
    ''' "range()" like function which accept float type''' 
    i = start
    while i < stop:
        yield i
        i += step
# Generate one element at a time.
# Preferred when you don't need all generated elements at the same time. 
# This will save memory.
for i in frange(1.0, 2.0, 0.5):
    print i   # Use generated element.
# Generate all elements at once.
# Preferred when generated list ought to be small.
print list(frange(1.0, 10.0, 0.5))    

输出:

1.0
1.5
[1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0, 4.5, 5.0, 5.5, 6.0, 6.5, 7.0, 7.5, 8.0, 8.5, 9.0, 9.5]