近似的数字 [英] aproximate a number
问题描述
大家好。我需要将给定的浮点数近似为下一个(int)
更大的浮点数。由于我的英语不好,我试着用一些例子来解释它:
5.7 - > 6
52.987 - > 53
3.34 - > 4
2.1 - > 3
问候
Hi all. I''d need to aproximate a given float number into the next (int)
bigger one. Because of my bad english I try to explain it with some example:
5.7 --> 6
52.987 --> 53
3.34 --> 4
2.1 --> 3
Regards
推荐答案
billiejoex写道:
billiejoex wrote:
大家好。我需要将给定的浮点数近似为下一个(int)
更大的浮点数。由于我的英语不好,我试着用一些例子来解释它:
5.7 - > 6
52.987 - > 53
3.34 - > 4
2.1 - > 3
Hi all. I''d need to aproximate a given float number into the next (int)
bigger one. Because of my bad english I try to explain it with some example:
5.7 --> 6
52.987 --> 53
3.34 --> 4
2.1 --> 3
Regards
math.ceil返回你需要的但是作为一个浮点数,然后创建一个int
math.ceil returns what you need but as a float, then create an int
import math
math.ceil(12.3)
13.0 int(math.ceil(12.3))
import math
math.ceil (12.3) 13.0 int (math.ceil (12.3))
13
hth
-
rafi
想象力比知识更重要。
(阿尔伯特爱因斯坦)
13
hth
--
rafi
"Imagination is more important than knowledge."
(Albert Einstein)
billiejoex写道:
billiejoex wrote:
大家好。我需要将给定的浮点数近似为下一个(int)
更大的浮点数。由于我的英语不好,我试着用一些例子来解释它:
5.7 - > 6
52.987 - > 53
3.34 - > 4
2.1 - > 3
Hi all. I''d need to aproximate a given float number into the next (int)
bigger one. Because of my bad english I try to explain it with some example:
5.7 --> 6
52.987 --> 53
3.34 --> 4
2.1 --> 3
看看math.ceil
Have a look at math.ceil
导入数学
math.ceil(5.7)
import math
math.ceil(5.7)
将McGugan
-
http://www.willmcgugan.com
"" .join({''*'':''@'',''^'':''。''}。get(c,0)或chr(97+ (ord(c)-84)%26)for c in
" jvyy * jvyyzpthtna ^ pbz")
6.0
Will McGugan
--
http://www.willmcgugan.com
"".join({''*'':''@'',''^'':''.''}.get(c,0) or chr(97+(ord(c)-84)%26) for c in
"jvyy*jvyyzpthtna^pbz")
billiejoex写道:
billiejoex wrote:
大家好。我需要将给定的浮点数近似为下一个(int)
更大的浮点数。由于我的英语不好,我试着用一些例子来解释它:
5.7 - > 6
52.987 - > 53
3.34 - > 4
2.1 - > 3
Hi all. I''d need to aproximate a given float number into the next (int)
bigger one. Because of my bad english I try to explain it with some example:
5.7 --> 6
52.987 --> 53
3.34 --> 4
2.1 --> 3
可能类似于int(数字+ 0.99999999),具体取决于你想要的
边界情况(你没有提到过)从技术上讲,
它应该是int(数字+ 1.0 - epsilon)。
-
Erik Max Francis& ;& ma*@alcyone.com && http://www.alcyone.com/max/
美国加利福尼亚州圣何塞&& 37 20 N 121 53 W& AIM erikmaxfrancis
如果你害怕寂寞,不要结婚。
- Anton Chekhov
Probably something like int(number + 0.99999999), depending on the
boundary cases you want (which you haven''t mentioned here. Technically,
it should be int(number + 1.0 - epsilon).
--
Erik Max Francis && ma*@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 20 N 121 53 W && AIM erikmaxfrancis
If you are afraid of loneliness, do not marry.
-- Anton Chekhov
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