from collections import Counter text = "d5851ijh7w6c3gb{hs1kso54he3{c5f4dsac{w9awdb2sd4gs{tefgtkk9b1917c9ehko41dtfc3ob2tgfc4557hsjj2oaagt{8{wh62cw1{i{f6hjhfl6gahwdwlf3bhtot}3w4b6dai47cw6o85{1haa2fctfcwdcfh8aolw{2al3{tie2jbead1o{49eesfw5f3oablhejls7hoaw5e1i1ls6c3c{5og6ojiswbsot3efs{d2j9o1d26ohecf395tkbhwfc2e8ciae2sfsbdaoc28iowf{ftw6cs1ftai141h4{o5ogwcastt{gkcg6w2hwofw88wtg145{4cjhh21hs3tadse{7t2{j2fecd7befbkhk1ogtds4kbsohls3hfadcgt44c2gci597ww66djit21j3hatjt4{fii7ht27cit1tswb{8c3kg9ht51bc27ss5o3fg7e3{sffebs7}5d7aw1sos63c3bts86h}9kseco36d8cfc4bs}bawb3t{1gf22ecicahgsa1{lh8}o4t1td3tdofbfafta{ow1d2}4ob{tfbc0000000" counter = Counter(text) sorted_counter = sorted(counter.items(), key=lambda x: x[1], reverse=True) for char, count in sorted_counter: print(f"{char}: {count}") 帮我修改一下这个脚本,使排序后的文本按照出现频率从高到低排列成一个字符串。
时间: 2023-11-01 13:44:09 浏览: 152
你可以使用列表推导式来实现这个目标。以下是修改后的代码:
```python
from collections import Counter
text = "d5851ijh7w6c3gb{hs1kso54he3{c5f4dsac{w9awdb2sd4gs{tefgtkk9b1917c9ehko41dtfc3ob2tgfc4557hsjj2oaagt{8{wh62cw1{i{f6hjhfl6gahwdwlf3bhtot}3w4b6dai47cw6o85{1haa2fctfcwdcfh8aolw{2al3{tie2jbead1o{49eesfw5f3oablhejls7hoaw5e1i1ls6c3c{5og6ojiswbsot3efs{d2j9o1d26ohecf395tkbhwfc2e8ciae2sfsbdaoc28iowf{ftw6cs1ftai141h4{o5ogwcastt{gkcg6w2hwofw88wtg145{4cjhh21hs3tadse{7t2{j2fecd7befbkhk1ogtds4kbsohls3hfadcgt44c2gci597ww66djit21j3hatjt4{fii7ht27cit1tswb{8c3kg9ht51bc27ss5o3fg7e3{sffebs7}5d7aw1sos63c3bts86h}9kseco36d8cfc4bs}bawb3t{1gf22ecicahgsa1{lh8}o4t1td3tdofbfafta{ow1d2}4ob{tfbc0000000"
counter = Counter(text)
sorted_counter = sorted(counter.items(), key=lambda x: x[1], reverse=True)
sorted_str = ''.join([char * count for char, count in sorted_counter])
print(sorted_str)
```
这段代码将会根据字符出现的频率从高到低排列字符串,并将结果打印出来。
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