blocks|key|3978187|text|您可以使用heapq.nlargest|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3978188|from+heapq+import+nlargest
for+score,+name+in+nlargest(3,+zip(scores,+names)):
++++print(name,+score)|code-block|syntax|javascript|3978189|这将输出以下内容：|3978190|Spec+450
Meng+300
Sparc+200|3978191|entityMap^0|5|E|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

You can use <code>heapq.nlargest</code>:

<pre><code>from heapq import nlargest
for score, name in nlargest(3, zip(scores, names)):
 print(name, score)
</code></pre>

This outputs:

<pre><code>Spec 450
Meng 300
Sparc 200
</code></pre>

blocks|key|3981307|text|如果您的问题出在打印部分：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3981308|for+i+in+range(3):
++++print(names[i],+scores[i])|code-block|syntax|javascript|3981309|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If your problem is in the print part:

<pre><code>for i in range(3):
 print(names[i], scores[i])
</code></pre>

blocks|key|3978085|text|你在最后一行的印刷是错误的。你不能像那样把list对象和str连接起来。改为执行以下操作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3978086|scores,+names+=+(list(t)+for+t+in+zip(*sorted(zip(scores,+names),+reverse=True)))

for+i+in+range(3):
++++print(names[i],+scores[i])|code-block|syntax|javascript|3978087|注意：reverse=True用于按逆序对列表进行排序。|3978088|entityMap^0|L|4|S|3|0|0|3|C|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

Your print in the last line is wrong. you can not concatenate <code>list</code> object with <code>str</code> like that. do this instead:

<pre><code>scores, names = (list(t) for t in zip(*sorted(zip(scores, names), reverse=True)))

for i in range(3):
 print(names[i], scores[i])
</code></pre>

NOTE: <code>reverse=True</code> is for sort the lists in reverse order (DESC).

blocks|key|3978146|text|看起来您正在尝试在print语句中连接两个列表和一个字符串。丹尼尔·莫勒的答案是正确的。此外，您可能需要将True传递给reverse=关键字参数，否则将得到最小的三个：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3978147|names+=+['Xander',+'Spec',+'Meng',+'Sparc',+'Jones',+'Nick',+'Link']
scores+=+[120,+450,+300,+200,+66,+183,+80]

scores,+names+=+(list(t)+for+t+in+zip(*sorted(zip(scores,+names),+reverse=True)))

for+i+in+range(3):
++++print(names[i],+scores[i])+

Spec+450
Meng+300
Sparc+200|code-block|syntax|javascript|3978148|entityMap^0|1H|4|1O|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

It looks like you're trying to concatenate two lists and a string within your print statement. Daniel Möller's answer is correct. Also you may need to pass <code>True</code> to the <code>reverse=</code> keyword argument, otherwise you'll get the smallest three:

<pre><code>names = ['Xander', 'Spec', 'Meng', 'Sparc', 'Jones', 'Nick', 'Link']
scores = [120, 450, 300, 200, 66, 183, 80]

scores, names = (list(t) for t in zip(*sorted(zip(scores, names), reverse=True)))

for i in range(3):
 print(names[i], scores[i]) 

Spec 450
Meng 300
Sparc 200
</code></pre>

blocks|key|3978169|text|以下是解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3978170|names+=+['Xander',+'Spec',+'Meng',+'Sparc',+'Jones',+'Nick',+'Link']
scores+=+[120,+450,+300,+200,+66,+183,+80]
data+=+list(zip(names,+scores))
print(data)
#+output+[('Xander',+120),+('Spec',+450),+('Meng',+300),+('Sparc',+200),+('Jones',+66),+('Nick',+183),+('Link',+80)]
data.sort(key=lambda+x:x[1],+reverse=True)
print(data)
#+output+[('Spec',+450),+('Meng',+300),+('Sparc',+200),+('Nick',+183),+('Xander',+120),+('Link',+80),+('Jones',+66)]


for+name,+score+in+data:
++++print(name,score,sep='\t')

"""+output+
Spec++++450
Meng++++300
Sparc+++200
Nick++++183
Xander++120
Link++++80
Jones+++66
"""|code-block|syntax|javascript|3978171|正如更改所暗示的那样，这可以以一种更优雅的方式来完成，如下所示。|3978172|names+=+['Xander',+'Spec',+'Meng',+'Sparc',+'Jones',+'Nick',+'Link']
scores+=+[120,+450,+300,+200,+66,+183,+80]
data+=+list(zip(scores,+names))
print(data)
#+output+[(120,+'Xander'),+(450,+'Spec'),+(300,+'Meng'),+(200,+'Sparc'),+(66,+'Jones'),+(183,+'Nick'),+(80,+'Link')]
data.sort(reverse=True)
print(data)
#+output+[(450,+'Spec'),+(300,+'Meng'),+(200,+'Sparc'),+(183,+'Nick'),+(120,+'Xander'),+(80,+'Link'),+(66,+'Jones')]

for+score,+name+in+data:
++++print(name,score,sep='\t')

"""+output+
Spec++++450
Meng++++300
Sparc+++200
Nick++++183
Xander++120
Link++++80
Jones+++66
"""|3978173|按照@Jab的建议使用itemgetter|3978174|names+=+['Xander',+'Spec',+'Meng',+'Sparc',+'Jones',+'Nick',+'Link']
scores+=+[120,+450,+300,+200,+66,+183,+80]
data+=+list(zip(names,+scores))
print(data)
#+output+[('Xander',+120),+('Spec',+450),+('Meng',+300),+('Sparc',+200),+('Jones',+66),+('Nick',+183),+('Link',+80)]
from+operator+import+itemgetter
data.sort(key=itemgetter(1),+reverse=True)
print(data)
#+output+[('Spec',+450),+('Meng',+300),+('Sparc',+200),+('Nick',+183),+('Xander',+120),+('Link',+80),+('Jones',+66)]


for+name,+score+in+data:
++++print(name,score,sep='\t')

"""+output+
Spec++++450
Meng++++300
Sparc+++200
Nick++++183
Xander++120
Link++++80
Jones+++66
"""|3978175|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

here is the solution 

<pre><code>names = ['Xander', 'Spec', 'Meng', 'Sparc', 'Jones', 'Nick', 'Link']
scores = [120, 450, 300, 200, 66, 183, 80]
data = list(zip(names, scores))
print(data)
# output [('Xander', 120), ('Spec', 450), ('Meng', 300), ('Sparc', 200), ('Jones', 66), ('Nick', 183), ('Link', 80)]
data.sort(key=lambda x:x[1], reverse=True)
print(data)
# output [('Spec', 450), ('Meng', 300), ('Sparc', 200), ('Nick', 183), ('Xander', 120), ('Link', 80), ('Jones', 66)]


for name, score in data:
 print(name,score,sep='\t')

""" output 
Spec 450
Meng 300
Sparc 200
Nick 183
Xander 120
Link 80
Jones 66
"""
</code></pre>

as changes suggested, this can be done in a more elegant way like this.

<pre><code>names = ['Xander', 'Spec', 'Meng', 'Sparc', 'Jones', 'Nick', 'Link']
scores = [120, 450, 300, 200, 66, 183, 80]
data = list(zip(scores, names))
print(data)
# output [(120, 'Xander'), (450, 'Spec'), (300, 'Meng'), (200, 'Sparc'), (66, 'Jones'), (183, 'Nick'), (80, 'Link')]
data.sort(reverse=True)
print(data)
# output [(450, 'Spec'), (300, 'Meng'), (200, 'Sparc'), (183, 'Nick'), (120, 'Xander'), (80, 'Link'), (66, 'Jones')]

for score, name in data:
 print(name,score,sep='\t')

""" output 
Spec 450
Meng 300
Sparc 200
Nick 183
Xander 120
Link 80
Jones 66
"""
</code></pre>

using itemgetter as suggested by @Jab 

<pre><code>names = ['Xander', 'Spec', 'Meng', 'Sparc', 'Jones', 'Nick', 'Link']
scores = [120, 450, 300, 200, 66, 183, 80]
data = list(zip(names, scores))
print(data)
# output [('Xander', 120), ('Spec', 450), ('Meng', 300), ('Sparc', 200), ('Jones', 66), ('Nick', 183), ('Link', 80)]
from operator import itemgetter
data.sort(key=itemgetter(1), reverse=True)
print(data)
# output [('Spec', 450), ('Meng', 300), ('Sparc', 200), ('Nick', 183), ('Xander', 120), ('Link', 80), ('Jones', 66)]


for name, score in data:
 print(name,score,sep='\t')

""" output 
Spec 450
Meng 300
Sparc 200
Nick 183
Xander 120
Link 80
Jones 66
"""
</code></pre>

blocks|key|3978219|text|您可以使用operator.itemgetter对数据进行压缩和排序，然后按如下方式打印数据：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3978220|from+operator+import+itemgetter

names+=+['Xander',+'Spec',+'Meng',+'Sparc',+'Jones',+'Nick',+'Link']
scores+=+[120,+450,+300,+200,+66,+183,+80]

data+=+sorted(zip(names,+scores),+key=itemgetter(1),+reverse=True)
for+name,+score+in+data[:3]:
++++print(f"{name}:+{score}")|code-block|syntax|javascript|3978221|这将输出以下内容：|3978222|Spec:+450
Meng:+300
Sparc:+200|3978223|entityMap^0|5|J|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

You can zip and sort the data using <code>operator.itemgetter</code> then print the data like so:

<pre><code>from operator import itemgetter

names = ['Xander', 'Spec', 'Meng', 'Sparc', 'Jones', 'Nick', 'Link']
scores = [120, 450, 300, 200, 66, 183, 80]

data = sorted(zip(names, scores), key=itemgetter(1), reverse=True)
for name, score in data[:3]:
 print(f"{name}: {score}")
</code></pre>

This outputs:

<pre><code>Spec: 450
Meng: 300
Sparc: 200
</code></pre>

blocks|key|3981756|text|如果names是唯一的，我建议使用dict，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3981757|names+=+['Xander',+'Spec',+'Meng',+'Sparc',+'Jones',+'Nick',+'Link']
scores+=+[120,+450,+300,+200,+66,+183,+80]
high+=+sorted([i+for+i+in+d.items()],key=lambda+x:x[1],reverse=True)[:3]
for+i+in+high:
++++print(i[0],i[1])|code-block|syntax|javascript|3981758|输出：|3981759|Spec+450
Meng+300
Sparc+200|3981760|sorted中的key用于指示它应该按值排序(第二个值，因为索引从0开始)，而reverse表示降序。|3981761|entityMap^0|2|5|H|4|0|0|0|0|0|6|8|3|13|7|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$]]

If <code>names</code> are unique I suggest usage of <code>dict</code>, following way:

<pre><code>names = ['Xander', 'Spec', 'Meng', 'Sparc', 'Jones', 'Nick', 'Link']
scores = [120, 450, 300, 200, 66, 183, 80]
high = sorted([i for i in d.items()],key=lambda x:x[1],reverse=True)[:3]
for i in high:
 print(i[0],i[1])
</code></pre>

Output:

<pre><code>Spec 450
Meng 300
Sparc 200
</code></pre>

<code>key</code> in <code>sorted</code> is used to point that it should sort by value (2nd one, as indexing start at 0) and <code>reverse</code> is meaning desceding order.

I'm having some problems with my code below. I have my two lists names and scores. These lists, correspond with each other as seen below. My goal is to print out the first three greatest items in both lists. I've attempted to sort them together from greatest to least and then print out the first three items, but I'm getting some unbounderror. Any thoughts? Thanks. 

<pre><code>names = ['Xander', 'Spec', 'Meng', 'Sparc', 'Jones', 'Nick', 'Link']
scores = [120, 450, 300, 200, 66, 183, 80]

scores, names = (list(t) for t in zip(*sorted(zip(scores, names))))
print(names[:3] + " " + scores[:3])
</code></pre>

Example Output:

<pre><code>Spec 450
Meng 300
Sparc 200
</code></pre>

Python sorting two lists together?

我在下面的代码中遇到一些问题。我有两份名单，名字和分数。这些列表彼此对应，如下所示。我的目标是打印出两个列表中最好的前三个项目。我尝试将它们从最大到最小排序在一起，然后打印出前三项，但我得到了一些无边界错误。有什么想法吗？谢谢。names = ['Xander', 'Spec', 'Meng', 'Sparc', 'Jones', 'Nick', 'Link']scores = [120, 450

问Python将两个列表排序在一起？
EN

回答 7

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Python将两个列表排序在一起？EN

回答 7

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Python将两个列表排序在一起？
EN