Python 如何使用十进制范围()步长值?
有没有办法在0和1之间按0.1步进 我以为我可以像下面这样做,但失败了:Python 如何使用十进制范围()步长值?,python,floating-point,range,Python,Floating Point,Range,有没有办法在0和1之间按0.1步进 我以为我可以像下面这样做,但失败了: for i in range(0, 1, 0.1): print i 相反,它说步长参数不能为零,这是我没有预料到的。增加循环的I的大小,然后在需要时减小它 for i * 100 in range(0, 100, 10): print i / 100.0 编辑:我真的不记得为什么我认为这会在语法上起作用 for i in range(0, 11, 1): print i / 10.0 它应该
for i in range(0, 1, 0.1):
print i
相反,它说步长参数不能为零,这是我没有预料到的。增加循环的
I
的大小,然后在需要时减小它
for i * 100 in range(0, 100, 10):
print i / 100.0
编辑:我真的不记得为什么我认为这会在语法上起作用
for i in range(0, 11, 1):
print i / 10.0
它应该具有所需的输出 恐怕range()内置函数会返回一系列整数值,因此不能用它来执行小数步
我要说的是使用while循环:
i = 0.0
while i <= 1.0:
print i
i += 0.1
i=0.0
而Python的range()只能做整数,不能做浮点。在您的具体情况下,您可以使用列表理解:
[x * 0.1 for x in range(0, 10)]
(将对范围的调用替换为该表达式。)
对于更一般的情况,您可能需要编写自定义函数或生成器。如果经常这样做,您可能需要保存生成的列表r
r=map(lambda x: x/10.0,range(0,10))
for i in r:
print i
在此基础上,您可以定义一个生成器,该生成器接受并生成您选择的任何类型(坚持支持+
和>>def drange(启动、停止、步骤)的类型):
…r=开始
…当r<停止时:
…收益率r
…r+=阶跃
...
>>>i0=drange(0.0,1.0,0.1)
>>>[%g”%x代表i0中的x]
['0', '0.1', '0.2', '0.3', '0.4', '0.5', '0.6', '0.7', '0.8', '0.9', '1']
>>>
与其直接使用小数步,不如用所需点数来表示。否则,浮点舍入误差可能会给出错误的结果
您可以使用库中的函数(它不是标准库的一部分,但相对容易获得)。linspace
需要返回许多点,还允许您指定是否包含正确的端点:
>>> np.linspace(0,1,11)
array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1. ])
>>> np.linspace(0,1,10,endpoint=False)
array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9])
如果您真的想使用浮点步骤值,可以使用numpy.arange
>>> import numpy as np
>>> np.arange(0.0, 1.0, 0.1)
array([ 0. , 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9])
不过,浮点舍入错误会导致问题。下面是一个简单的例子,舍入错误会导致arange
生成一个长度为4的数组,而它应该只生成3个数字:
>>> numpy.arange(1, 1.3, 0.1)
array([1. , 1.1, 1.2, 1.3])
您可以使用此功能:
def frange(start,end,step):
return map(lambda x: x*step, range(int(start*1./step),int(end*1./step)))
为可能出现的错误登录步骤添加自动更正:
def frange(start,step,stop):
step *= 2*((stop>start)^(step<0))-1
return [start+i*step for i in range(int((stop-start)/step))]
def frange(启动、步骤、停止):
步骤*=2*((停止>开始)^(步骤
在Python 2.7x中,您可以得到以下结果:
[0.0,0.1,0.2,0.300000000004,0.4,0.5,0.600000000000001,0.70000000000001,0.8,0.9]
但如果您使用:
[ round(x * 0.1, 1) for x in range(0, 10)]
为您提供所需的:
[0.0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9]
与seq
函数类似,该函数返回给定正确步长值的任意顺序的序列。最后一个值等于停止值
def seq(start, stop, step=1):
n = int(round((stop - start)/float(step)))
if n > 1:
return([start + step*i for i in range(n+1)])
elif n == 1:
return([start])
else:
return([])
结果
[1.0,1.5,2.0,2.5,3.0,3.5,4.0,4.5,5.0]
[10,9,8,7,6,5,4,3,2,1,0]
[10,8,6,4,2,0]
[1]
我的解决方案:
def seq(start, stop, step=1, digit=0):
x = float(start)
v = []
while x <= stop:
v.append(round(x,digit))
x += step
return v
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)
def seq(开始、停止、步骤=1,数字=0):
x=浮动(开始)
v=[]
而x这里有一个解决方案,使用:
用法示例:
for i in seq(0, 1, 0.1):
print(i)
我的版本使用原始范围函数为移位创建乘法索引。这允许使用与原始范围函数相同的语法。
我制作了两个版本,一个使用浮点,一个使用十进制,因为我发现在某些情况下,我希望避免浮点运算引入的舍入漂移
它与range/xrange中的空集结果一致
只向任一函数传递一个数值将使标准范围输出返回到输入参数的整数上限值(因此,如果给它5.5,它将返回范围(6))
编辑:下面的代码现在作为pypi上的软件包提供:
下面是我的解决方案,它可以很好地处理浮点值范围(-1,0,0.01),并且没有浮点表示错误。它不是很快,但可以很好地工作:
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]
从十进制导入十进制
def get_乘数(_from,_to,step):
数字=[]
对于[_-from,_-to,step]中的数字:
pre=十进制(str(number))%1
数字=len(str(pre))-2
数字。追加(数字)
最大位数=最大(位数)
返回浮点数(10**(最大位数))
def float_范围(_from,_to,step,include=False):
“”“在范围[开始,停止]上生成浮点值的范围列表”
步长
include=True-如果可能,允许将正确的值包含到
!!适用于浮点表示!!
"""
mult=获取乘法器(_-from,_-to,step)
#打印mult
int_from=int(圆形(_from*mult))
int_to=int(圆形(_to*mult))
整数步=整数(四舍五入(步*mult))
#打印整数从,整数到,整数步
如果包括:
结果=范围(整数从,整数到+整数步长,整数步长)
result=[r for r in result if r我只是一个初学者,但在模拟一些计算时,我遇到了同样的问题。下面是我如何尝试解决这个问题的,这似乎是在使用十进制步骤
我也很懒,所以我发现很难编写自己的范围函数
基本上,我所做的是将我的xrange(0.0,1.0,0.01)
更改为xrange(0,100,1)
,并在循环中使用除以100.0
。
我还担心是否会出现舍入错误。因此我决定测试是否存在舍入错误。现在我听说,如果计算中的例如0.01
不完全是浮点0.01
比较它们应该返回False(如果我错了,请让我知道)
因此,我决定通过运行一个简短的测试来测试我的解决方案是否适用于我的范围:
for d100 in xrange(0, 100, 1):
d = d100 / 100.0
fl = float("0.00"[:4 - len(str(d100))] + str(d100))
print d, "=", fl , d == fl
每一个都是这样
现在,如果我完全弄错了,请让我知道。这是我的解决方案,可以通过浮动步长获得范围。
使用此函数,无需导入或安装numpy。
我是
seq(1, 1)
def seq(start, stop, step=1, digit=0):
x = float(start)
v = []
while x <= stop:
v.append(round(x,digit))
x += step
return v
import numpy as np
for i in np.arange(0, 1, 0.1):
print i
import itertools
def seq(start, end, step):
if step == 0:
raise ValueError("step must not be 0")
sample_count = int(abs(end - start) / step)
return itertools.islice(itertools.count(start, step), sample_count)
for i in seq(0, 1, 0.1):
print(i)
## frange.py
from math import ceil
# find best range function available to version (2.7.x / 3.x.x)
try:
_xrange = xrange
except NameError:
_xrange = range
def frange(start, stop = None, step = 1):
"""frange generates a set of floating point values over the
range [start, stop) with step size step
frange([start,] stop [, step ])"""
if stop is None:
for x in _xrange(int(ceil(start))):
yield x
else:
# create a generator expression for the index values
indices = (i for i in _xrange(0, int((stop-start)/step)))
# yield results
for i in indices:
yield start + step*i
## drange.py
import decimal
from math import ceil
# find best range function available to version (2.7.x / 3.x.x)
try:
_xrange = xrange
except NameError:
_xrange = range
def drange(start, stop = None, step = 1, precision = None):
"""drange generates a set of Decimal values over the
range [start, stop) with step size step
drange([start,] stop, [step [,precision]])"""
if stop is None:
for x in _xrange(int(ceil(start))):
yield x
else:
# find precision
if precision is not None:
decimal.getcontext().prec = precision
# convert values to decimals
start = decimal.Decimal(start)
stop = decimal.Decimal(stop)
step = decimal.Decimal(step)
# create a generator expression for the index values
indices = (
i for i in _xrange(
0,
((stop-start)/step).to_integral_value()
)
)
# yield results
for i in indices:
yield float(start + step*i)
## testranges.py
import frange
import drange
list(frange.frange(0, 2, 0.5)) # [0.0, 0.5, 1.0, 1.5]
list(drange.drange(0, 2, 0.5, precision = 6)) # [0.0, 0.5, 1.0, 1.5]
list(frange.frange(3)) # [0, 1, 2]
list(frange.frange(3.5)) # [0, 1, 2, 3]
list(frange.frange(0,10, -1)) # []
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]
for d100 in xrange(0, 100, 1):
d = d100 / 100.0
fl = float("0.00"[:4 - len(str(d100))] + str(d100))
print d, "=", fl , d == fl
from __future__ import division
from math import log
def xfrange(start, stop, step):
old_start = start #backup this value
digits = int(round(log(10000, 10)))+1 #get number of digits
magnitude = 10**digits
stop = int(magnitude * stop) #convert from
step = int(magnitude * step) #0.1 to 10 (e.g.)
if start == 0:
start = 10**(digits-1)
else:
start = 10**(digits)*start
data = [] #create array
#calc number of iterations
end_loop = int((stop-start)//step)
if old_start == 0:
end_loop += 1
acc = start
for i in xrange(0, end_loop):
data.append(acc/magnitude)
acc += step
return data
print xfrange(1, 2.1, 0.1)
print xfrange(0, 1.1, 0.1)
print xfrange(-1, 0.1, 0.1)
[1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0]
[0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1]
[-1.0, -0.9, -0.8, -0.7, -0.6, -0.5, -0.4, -0.3, -0.2, -0.1, 0.0]
# floating point range
def frange(a, b, stp=1.0):
i = a+stp/2.0
while i<b:
yield a
a += stp
i += stp
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)
for i in np.arange(0, 1, 0.1).tolist():
print i
>>> step = .1
>>> N = 10 # number of data points
>>> [ x / pow(step, -1) for x in range(0, N + 1) ]
[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]
>>> step = .1
>>> rnge = 1 # NOTE range = 1, i.e. span of data points
>>> N = int(rnge / step
>>> [ x / pow(step,-1) for x in range(0, N + 1) ]
[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0]
>>> import math
>>> def f(x):
return math.sin(x)
>>> step = .1
>>> rnge = 1 # NOTE range = 1, i.e. span of data points
>>> N = int(rnge / step)
>>> [ f( x / pow(step,-1) ) for x in range(0, N + 1) ]
[0.0, 0.09983341664682815, 0.19866933079506122, 0.29552020666133955, 0.3894183423086505,
0.479425538604203, 0.5646424733950354, 0.644217687237691, 0.7173560908995228,
0.7833269096274834, 0.8414709848078965]
def xdt(n):
return dt*float(n)
tlist = map(xdt, range(int(t_max/dt)+1))
[p/10 for p in range(0, 10)]
[0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9]
x=100
y=2
[p/x for p in range(0, int(x*y))]
[0.0, 0.01, 0.02, 0.03, ..., 1.97, 1.98, 1.99]
import more_itertools as mit
for x in mit.numeric_range(0, 1, 0.1):
print("{:.1f}".format(x))
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
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")))
print(linspace(0, 10, 5))
# linspace(0, 10, 5)
print(list(linspace(0, 10, 5)))
# [0.0, 2.5, 5.0, 7.5, 10]
import collections.abc
import numbers
class linspace(collections.abc.Sequence):
"""linspace(start, stop, num) -> linspace object
Return a virtual sequence of num numbers from start to stop (inclusive).
If you need a half-open range, use linspace(start, stop, num+1)[:-1].
"""
def __init__(self, start, stop, num):
if not isinstance(num, numbers.Integral) or num <= 1:
raise ValueError('num must be an integer > 1')
self.start, self.stop, self.num = start, stop, num
self.step = (stop-start)/(num-1)
def __len__(self):
return self.num
def __getitem__(self, i):
if isinstance(i, slice):
return [self[x] for x in range(*i.indices(len(self)))]
if i < 0:
i = self.num + i
if i >= self.num:
raise IndexError('linspace object index out of range')
if i == self.num-1:
return self.stop
return self.start + i*self.step
def __repr__(self):
return '{}({}, {}, {})'.format(type(self).__name__,
self.start, self.stop, self.num)
def __eq__(self, other):
if not isinstance(other, linspace):
return False
return ((self.start, self.stop, self.num) ==
(other.start, other.stop, other.num))
def __ne__(self, other):
return not self==other
def __hash__(self):
return hash((type(self), self.start, self.stop, self.num))
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
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')]
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
def frange(x, y, step):
if int(x + y + step) == (x + y + step):
r = list(range(int(x), int(y), int(step)))
else:
f = 10 ** (len(str(step)) - str(step).find('.') - 1)
rf = list(range(int(x * f), int(y * f), int(step * f)))
r = [i / f for i in rf]
return r