我如何在0和1之间迭代0.1步?
这表示step参数不能为零:
for i in range(0, 1, 0.1):
print(i)
我如何在0和1之间迭代0.1步?
这表示step参数不能为零:
for i in range(0, 1, 0.1):
print(i)
当前回答
sign = lambda x: (1, -1)[x < 0]
def frange(start, stop, step):
i = 0
r=len(str(step).split('.')[-1])
args=(start,stop,step)
if not step :return []
if all(int(i)==float(i) for i in args):
start,stop,step=map(int,args)
if sign(step)==1:
while start + i * step < stop:
yield round(start + i * step,r)
i += 1
else:
while start + i * step > stop:
yield round(start + i * step,r)
i += 1
其他回答
可以使用Numpy库来完成。Arange()函数允许浮动中的步骤。但是,它返回一个numpy数组,为了方便起见,可以使用tolist()将其转换为列表。
for i in np.arange(0, 1, 0.1).tolist():
print i
为了解决浮点精度问题,可以使用Decimal模块。
这要求在编写代码时将int或float转换为Decimal,但如果确实需要这种便利,则可以传递str并修改函数。
from decimal import Decimal
def decimal_range(*args):
zero, one = Decimal('0'), Decimal('1')
if len(args) == 1:
start, stop, step = zero, args[0], one
elif len(args) == 2:
start, stop, step = args + (one,)
elif len(args) == 3:
start, stop, step = args
else:
raise ValueError('Expected 1 or 2 arguments, got %s' % len(args))
if not all([type(arg) == Decimal for arg in (start, stop, step)]):
raise ValueError('Arguments must be passed as <type: Decimal>')
# neglect bad cases
if (start == stop) or (start > stop and step >= zero) or \
(start < stop and step <= zero):
return []
current = start
while abs(current) < abs(stop):
yield current
current += step
输出样本-
from decimal import Decimal as D
list(decimal_range(D('2')))
# [Decimal('0'), Decimal('1')]
list(decimal_range(D('2'), D('4.5')))
# [Decimal('2'), Decimal('3'), Decimal('4')]
list(decimal_range(D('2'), D('4.5'), D('0.5')))
# [Decimal('2'), Decimal('2.5'), Decimal('3.0'), Decimal('3.5'), Decimal('4.0')]
list(decimal_range(D('2'), D('4.5'), D('-0.5')))
# []
list(decimal_range(D('2'), D('-4.5'), D('-0.5')))
# [Decimal('2'),
# Decimal('1.5'),
# Decimal('1.0'),
# Decimal('0.5'),
# Decimal('0.0'),
# Decimal('-0.5'),
# Decimal('-1.0'),
# Decimal('-1.5'),
# Decimal('-2.0'),
# Decimal('-2.5'),
# Decimal('-3.0'),
# Decimal('-3.5'),
# Decimal('-4.0')]
如果你经常这样做,你可能想保存生成的列表r
r=map(lambda x: x/10.0,range(0,10))
for i in r:
print i
增加循环的i的大小,然后在需要时减少它。
for i * 100 in range(0, 100, 10):
print i / 100.0
编辑:老实说,我不记得为什么我认为这会在语法上工作
for i in range(0, 11, 1):
print i / 10.0
这应该有期望的输出。
下面是我的解决方案,它可以很好地使用float_range(- 1,0,0.01),并且没有浮点表示错误。它不是很快,但工作得很好:
from decimal import Decimal
def get_multiplier(_from, _to, step):
digits = []
for number in [_from, _to, step]:
pre = Decimal(str(number)) % 1
digit = len(str(pre)) - 2
digits.append(digit)
max_digits = max(digits)
return float(10 ** (max_digits))
def float_range(_from, _to, step, include=False):
"""Generates a range list of floating point values over the Range [start, stop]
with step size step
include=True - allows to include right value to if possible
!! Works fine with floating point representation !!
"""
mult = get_multiplier(_from, _to, step)
# print mult
int_from = int(round(_from * mult))
int_to = int(round(_to * mult))
int_step = int(round(step * mult))
# print int_from,int_to,int_step
if include:
result = range(int_from, int_to + int_step, int_step)
result = [r for r in result if r <= int_to]
else:
result = range(int_from, int_to, int_step)
# print result
float_result = [r / mult for r in result]
return float_result
print float_range(-1, 0, 0.01,include=False)
assert float_range(1.01, 2.06, 5.05 % 1, True) ==\
[1.01, 1.06, 1.11, 1.16, 1.21, 1.26, 1.31, 1.36, 1.41, 1.46, 1.51, 1.56, 1.61, 1.66, 1.71, 1.76, 1.81, 1.86, 1.91, 1.96, 2.01, 2.06]
assert float_range(1.01, 2.06, 5.05 % 1, False)==\
[1.01, 1.06, 1.11, 1.16, 1.21, 1.26, 1.31, 1.36, 1.41, 1.46, 1.51, 1.56, 1.61, 1.66, 1.71, 1.76, 1.81, 1.86, 1.91, 1.96, 2.01]