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<p>注意:我是Python新手。你知道吗</p>
<p>我的任务是设计一个程序,可以输出<strong>外国的</strong>车牌号,但是如果他们超速的话只能输出</strong>。我可能在这一过程中犯了一些错误,但我需要帮助列出计时2和字典计时3。
<code>#UK</code>和<code>#F</code>只是我的笔记,以便我能够快速查看哪个是英国牌照,哪个是外国牌照。你知道吗</p>
<pre><code>import re
distance=750 #variable for the distance between the Camera A and B (in m)
speedlimit=70 # (mps)
NumberPlates=["DV61 GGB",#UK
"DS11 EUBG 20",#F
"5T314 10A02",#F
"24TEG 5063",#F
"TR09 TRE",#UK
"524 WAL 75",#F
"TR44 VCZ",#UK
"FR52 SWD",#UK
"100 GBS 12",#F
"HG55 BPO"#UK
]
Enter=[7.12,7.15,7.24,7.45,7.28,7.31,7.18,7.25,7.33,7.38] #A list for the times of cars passing Camera A
Leave=[7.56,7.24,7.48,7.52,7.45,7.57,7.22,7.31,7.37,7.47] #A list for the times of cars passing Camera B
Timestaken=[]
Timestaken2=[]
Timestaken3={}
for enter_data, leave_data in zip(Enter, Leave):
Timestaken.append(leave_data-enter_data)
Timestaken=["%.2f" % (leave_data-enter_data) for enter_data, leave_data in zip(Enter, Leave)]
Timestaken2=[s.strip("0") for s in Timestaken]
Timestaken2=[s.strip('.') for s in Timestaken2]
for key,value in zip(NumberPlates,Timestaken2):
Timestaken3[key]=value
print(Timestaken3)
for item in NumberPlates:
UK_Numbers=list(filter(lambda x: re.match("[A-Z]{2}\d{2}\s+[A-Z]{3}$",x),NumberPlates))
for item in UK_Numbers:
if item in UK_Numbers:
NumberPlates.remove(item)
print(NumberPlates)
for key,value in zip(NumberPlates,Timestaken2):
Timestaken3[key]=value
print(Timestaken3)
print("10 cars have passed Camera A, then Camera B\n")
for key,value in Timestaken3.items():
speed=distance/int(value)
if speed>speedlimit:
print(key,"is speeding with",speed,"mps")
</code></pre>
<p>我在程序中加入了print(),以查看程序最终执行的操作。第二次是:</p>
<pre><code>for key,value in zip(NumberPlates,Timestaken2):
Timestaken3[key]=value
</code></pre>
<p>我只希望剩下的数字板,这将是外国的打印后,这个代码的权利。Timestaken2仍然有10个值,这是问题吗?你知道吗</p>
<p>请帮我解决任何问题。你知道吗</p>