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
我知道我在这里迟到了,但这里有一个简单的生成器解决方案,在3.6中工作:
def floatRange(*args):
start, step = 0, 1
if len(args) == 1:
stop = args[0]
elif len(args) == 2:
start, stop = args[0], args[1]
elif len(args) == 3:
start, stop, step = args[0], args[1], args[2]
else:
raise TypeError("floatRange accepts 1, 2, or 3 arguments. ({0} given)".format(len(args)))
for num in start, step, stop:
if not isinstance(num, (int, float)):
raise TypeError("floatRange only accepts float and integer arguments. ({0} : {1} given)".format(type(num), str(num)))
for x in range(int((stop-start)/step)):
yield start + (x * step)
return
然后你可以像原来的range()一样调用它…没有错误处理,但请告诉我是否有可以合理捕捉到的错误,我会更新。或者你可以更新它。这是StackOverflow。
其他回答
在'xrange([start], stop[, step])'基础上,你可以定义一个生成器,它接受并生成你选择的任何类型(坚持支持+和<的类型):
>>> def drange(start, stop, step):
... r = start
... while r < stop:
... yield r
... r += step
...
>>> i0=drange(0.0, 1.0, 0.1)
>>> ["%g" % x for x in i0]
['0', '0.1', '0.2', '0.3', '0.4', '0.5', '0.6', '0.7', '0.8', '0.9', '1']
>>>
为了精品的完整性,一个功能性的解决方案:
def frange(a,b,s):
return [] if s > 0 and a > b or s < 0 and a < b or s==0 else [a]+frange(a+s,b,s)
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
f = lambda x,y,z: (x+i*z for i in range(int((y-x)/z)))
上面不需要借助任何库就可以做分数步。
Start和stop是包含的,而不是一个或另一个(通常不包括stop),并且没有导入,并使用生成器
def rangef(start, stop, step, fround=5):
"""
Yields sequence of numbers from start (inclusive) to stop (inclusive)
by step (increment) with rounding set to n digits.
:param start: start of sequence
:param stop: end of sequence
:param step: int or float increment (e.g. 1 or 0.001)
:param fround: float rounding, n decimal places
:return:
"""
try:
i = 0
while stop >= start and step > 0:
if i==0:
yield start
elif start >= stop:
yield stop
elif start < stop:
if start == 0:
yield 0
if start != 0:
yield start
i += 1
start += step
start = round(start, fround)
else:
pass
except TypeError as e:
yield "type-error({})".format(e)
else:
pass
# passing
print(list(rangef(-100.0,10.0,1)))
print(list(rangef(-100,0,0.5)))
print(list(rangef(-1,1,0.2)))
print(list(rangef(-1,1,0.1)))
print(list(rangef(-1,1,0.05)))
print(list(rangef(-1,1,0.02)))
print(list(rangef(-1,1,0.01)))
print(list(rangef(-1,1,0.005)))
# failing: type-error:
print(list(rangef("1","10","1")))
print(list(rangef(1,10,"1")))
Python 3.6.2 (v3.6.2:5fd33b5, july 8 2017, 04:57:36) [MSC v.1900 64 . 位(AMD64)]