[英]Compare 2 strings with repetitive parts
I am developing a speak reconition trainer and I want to get the hits and errors of the trainer result. 我正在开发一个语音复习教练,我想知道教练结果的成败。 I have the real sentence and the predicted sentence and I have something like this: 我有真实的句子和预测的句子,并且我有这样的东西:
if real == predicted:
hits += 1
else:
errors += 1
My problem comes now... The predicted strings are with the form: 我的问题来了...预测的字符串具有以下形式:
- 'command' 'time' 'unit'
Where 'unit' can be [minute,minutes,second,seconds] And I want to count as hit the sentence with minute or minutes and second or secons. “单位”可以是[分钟,分钟,秒,秒],而我想算为命中分钟或分钟,秒或秒数的句子。 For example. 例如。
Real: stop five minutes and walk one second Predicted1: stop five minute and walk one second Predicted2: stop five minutes and walk one seconds Predicted3: stop three minute and walk one seconds 真实:停止五分钟,走一秒钟Predicted1:停止五分钟,走一秒钟Predicted2:停止五分钟,走一秒钟Predicted3:停止三分钟,走一秒钟
Where Predicted1 and Predicted2 are hits and Predicted3 an error. 命中Predicted1和Predicted2,Predicted3是错误。 There are a fast way without turning all as lists and this kind of things 有一种快速的方法可以不将所有内容都列出来
A quick solution would be to strip the rightmost s
from the units and then compare. 一种快速的解决方案是从单元中剥离最右边的s
,然后进行比较。
That is, if the unit is in variable unit
, then unit.rstrip('s')
returns the units without the rightmost s
, and doesn't change the unit if it didn't have an s
in the first place. 也就是说,如果该单元是在可变unit
,然后unit.rstrip('s')
返回单元而不最右边的s
,并且如果它不具有不改变单元s
在首位。
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