怎么统计这个文件中出现频率排前10的整数。（代码）
生成一个大文件data.txt,要求1200行，每行随即为0~20的整数。读取data.txt文件，统计这个文件中出现频率排前10的整数。
# 写入
import random
with open("data.txt", "w", encoding="utf-8") as f:
s = ""
for i in range(1, 1201):
s += str(random.randint(0, 20)) + "\n"
f.write(s[:-1])
# 读取
d = dict([(i, 0) for i in range(0, 21)])
with open("data.txt", "r", encoding="utf-8") as f:
n = f.read().split("\n")
for i in n:
d += 1
result = for i in sorted(d.items(), reverse=True, key=lambda d: d)][:10]
for i in result:
print(i) from operator import itemgetter
import random
# 生成文件
with open('data.txt', 'w') as fp:
lst = (str(random.randint(0,20))+'\n' for _ in range(1200))
fp.writelines(lst)
# 统计文件
with open('data.txt') as fp:
result = {}
for line in fp:
temp = int(line[:-1])
result = result.get(temp, 0) + 1
result = dict(sorted(result.items(), key=itemgetter(1,0), reverse=True))
for k, v in tuple(result.items())[:10]:
print(f'随机数：{k}, 出现次数：{v}') isdkz 发表于 2022-3-20 12:41
大佬{:5_92:} ckblt 发表于 2022-3-20 12:38
{:10_275:}
页:
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