python函数中的谐波意思是? [英] Harmonic mean in a python function?
问题描述
我有2个函数给出精确度和查全率,我需要在使用这两个分数的同一个库中定义一个谐波均值函数.函数看起来像这样:
I have 2 functions that give out precision and recall scores, I need to make a harmonic mean function defined in the same library that uses these two scores. The functions looks like this:
以下是功能:
def precision(ref, hyp):
"""Calculates precision.
Args:
- ref: a list of 0's and 1's extracted from a reference file
- hyp: a list of 0's and 1's extracted from a hypothesis file
Returns:
- A floating point number indicating the precision of the hypothesis
"""
(n, np, ntp) = (len(ref), 0.0, 0.0)
for i in range(n):
if bool(hyp[i]):
np += 1
if bool(ref[i]):
ntp += 1
return ntp/np
def recall(ref, hyp):
"""Calculates recall.
Args:
- ref: a list of 0's and 1's extracted from a reference file
- hyp: a list of 0's and 1's extracted from a hypothesis file
Returns:
- A floating point number indicating the recall rate of the hypothesis
"""
(n, nt, ntp) = (len(ref), 0.0, 0.0)
for i in range(n):
if bool(ref[i]):
nt += 1
if bool(hyp[i]):
ntp += 1
return ntp/nt
谐波均值函数是什么样子? 我所拥有的只是这个,但我知道这是不对的:
What would the harmonic mean function look like? All I have is this but I know its not right:
def F1(precision, recall):
(2*precision*recall)/(precision+recall)
推荐答案
对您的F1
函数稍作更改,并使用与您定义的相同的precision
和recall
函数,我可以正常工作:>
With a slight change of your F1
function, and with the same precision
and recall
function you defined, I have this working:
def F1(precision, recall):
return (2*precision*recall)/(precision+recall)
r = [0,1,0,0,0,1,1,0,1]
h = [0,1,1,1,0,0,1,0,1]
p = precision(r, h)
rec = recall(r, h)
f = F1(p, rec)
print f
特别回顾一下我拥有的变量的使用.您必须计算每个函数的结果并将其传递给F1
函数.
Review especially the use of variables I have. You must compute the result of each function and pass them to the F1
function.
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