blocks|key|4269419|text|更改此行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4269420|for+numbers+in+a:|code-block|syntax|javascript|4269421|至：|4269422|for+i,+numbers+in+enumerate(a):|4269423|然后更改打印索引的方式：|4269424|print("Number",+n,+"is+located+in+the+list+and+the+position+is:",+(i%2B1))|4269425|示例输出：|4269426|[1,+2,+2,+5,+5,+5,+6,+7,+8,+8,+8,+10,+10,+10,+10,+10,+11,+11,+16,+17,+17,+19,+19]
What+number+are+you+looking+for?+8
Number+8+is+located+in+the+list+and+the+position+is:+9
Number+8+is+located+in+the+list+and+the+position+is:+10
Number+8+is+located+in+the+list+and+the+position+is:+11|4269427|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Change this line:

<pre><code>for numbers in a:
</code></pre>

To:

<pre><code>for i, numbers in enumerate(a):
</code></pre>

And then change how you print the indices:

<pre><code>print("Number", n, "is located in the list and the position is:", (i+1))
</code></pre>

Sample output:

<pre><code>[1, 2, 2, 5, 5, 5, 6, 7, 8, 8, 8, 10, 10, 10, 10, 10, 11, 11, 16, 17, 17, 19, 19]
What number are you looking for? 8
Number 8 is located in the list and the position is: 9
Number 8 is located in the list and the position is: 10
Number 8 is located in the list and the position is: 11
</code></pre>

blocks|key|4269905|text|num+=20

numlist+=+[2,+4,+8,+9,+10,+10,+16,+19,+20,+20]

for+each+in+numlist:
++++if+num+is+each:
++++++++print+num
++++++++print+[i+for+i,+x+in+enumerate(numlist)+if+x+==+num]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4269906|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>num =20

numlist = [2, 4, 8, 9, 10, 10, 16, 19, 20, 20]

for each in numlist:
 if num is each:
 print num
 print [i for i, x in enumerate(numlist) if x == num]
</code></pre>

blocks|key|4269951|text|您可以使用numpy来删除循环，从而简化这段代码。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4269952|a+=+np.array([random.randint(1,20)+for+i+in+range(random.randint(8,30))])|code-block|syntax|javascript|4269953|然后，您可以使用np.where来确定用户是否在随机值数组a中选择了一个值：|4269954|idx_selections+=+np.where(a+==+n)[0]|4269955|然后，您可以处理用户是否匹配答案：|4269956|if+len(idx_selections)+==+0:
++++print("Your+number+could+not+be+found")
else:
++++for+i+in+idx_selections:
++++++++print("Number",+n,+"is+located+in+the+list+and+the+position+is:",+i)|4269957|entityMap^0|5|5|0|0|8|8|T|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|14|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

You could simplify this code using <code>numpy</code> to remove your loops.

<pre><code>a = np.array([random.randint(1,20) for i in range(random.randint(8,30))])
</code></pre>

Then you can use <code>np.where</code> to determine if the user selected a value in your array <code>a</code> of random values:

<pre><code>idx_selections = np.where(a == n)[0]
</code></pre>

Then you can handle if the user matched an answer or not:

<pre><code>if len(idx_selections) == 0:
 print("Your number could not be found")
else:
 for i in idx_selections:
 print("Number", n, "is located in the list and the position is:", i)
</code></pre>

blocks|key|4269566|text|如果你觉得很棒，你可以把你的一些循环转换成列表理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4269567|def+askUser():

++++n+=+input("What+number+are+you+looking+for?")
++++while+not+n.isdigit():
++++++++n+=+input("What+number+are+you+looking+for?")

++++n+=+int(n)

++++#+get+a+list+of+all+indexes+that+match+the+number
++++foundAt+=+[p%2B1+for+p,num+in+enumerate(a)+if+num+==+n]

++++if+foundAt:
++++++++#+print+text+by+creating+a+list+of+texts+to+print+and+decompose+them
++++++++#+printing+with+a+seperator+of+linefeed
++++++++print(+*[f"Number+{n}+is+located+in+the+list+and+the+position+is:+{q}"+for+
+++++++++++++++++q+in+foundAt],+sep="\n")
++++else:+
++++++++print("Your+number+could+not+be+found")|code-block|syntax|javascript|4269568|编辑:正如克里斯茨指出的那样，Python3.6的PEP-498附带了f""格式的字符串(不知道:o/+)-所以对于更早的3.x版本，Python必须使用|offset|length|style|CODE|4269569|print(+*["Number+{}+is+located+in+the+list+and+the+position+is:+{}".format(n,q)+for+
+++++++++++++++++q+in+foundAt],+sep="\n")|4269570|entityMap|0|LINK|mutability|MUTABLE|url|https://www.python.org/dev/peps/pep-0498/?^0|0|0|Z|3|P|7|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@$I|11|J|12|1|13]]|A|$]]|$1|M|3|N|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

If you feel fancy you can convert some of your loops into list comprehensions:

<pre><code>def askUser():

 n = input("What number are you looking for?")
 while not n.isdigit():
 n = input("What number are you looking for?")

 n = int(n)

 # get a list of all indexes that match the number
 foundAt = [p+1 for p,num in enumerate(a) if num == n]

 if foundAt:
 # print text by creating a list of texts to print and decompose them
 # printing with a seperator of linefeed
 print( *[f"Number {n} is located in the list and the position is: {q}" for 
 q in foundAt], sep="\n")
 else: 
 print("Your number could not be found")
</code></pre>

Edit: as Chrisz pointed out <code>f""</code> format strings came with <a href="https://www.python.org/dev/peps/pep-0498/?" rel="nofollow noreferrer">PEP-498</a> for Python 3.6 (wasn't aware :o/ ) - so for earlier 3.x Python would have to use 

<pre><code>print( *["Number {} is located in the list and the position is: {}".format(n,q) for 
 q in foundAt], sep="\n")
</code></pre>

So I have written a code that generates a random list with random number of values. Then asks the user what number the user is looking for and if its in the list, it will tell the user what position in the list the number is in.

<pre><code>import random

a = [random.randint(1, 20) for i in range(random.randint(8, 30))]

a.sort()
print(a)


def askUser():

 n = input("What number are you looking for?")
 while not n.isdigit():
 n = input("What number are you looking for?")

 n = int(n)
 s = 0

 for numbers in a:
 if numbers == n:
 s += 1
 print("Number", n, "is located in the list and the position is:", (a.index(n)+1))
 # Something here to skip this index next time it goes through the loop
 else:
 pass
 if s == 0:
 print("Your number could not be found")


askUser()
</code></pre>

I would like to add something that will skip the index it found first time and then look for the index of the duplicate if there is one.

Current result

<pre><code>[2, 4, 8, 9, 10, 10, 16, 19, 20, 20]
What number are you looking for?20
Number 20 is located in the list and the position is: 9
Number 20 is located in the list and the position is: 9
</code></pre>

Desired result

<pre><code>[2, 4, 8, 9, 10, 10, 16, 19, 20, 20]
What number are you looking for?20
Number 20 is located in the list and the position is: 9
Number 20 is located in the list and the position is: 10
</code></pre>

Checking duplicates in list from user input

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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所以我已经写了一段代码，它生成一个随机列表，其中包含随机数个值。然后询问用户正在寻找的数字，如果它在列表中，它将告诉用户该数字在列表中的位置。import randoma = [random.randint(1, 20) for i in range(random.randint(8, 30))]a.sort()pri...

问从用户输入检查列表中的重复项
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