[已解决]怎么统计这个文件中出现频率排前10的整数。（代码）

yangbaowen · 发表于 2022-3-20 12:13:09

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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。

最佳答案

月排行榜 / 总排行榜

isdkz

2022-3-20 12:41:11

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[temp] = 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}')
复制代码

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ckblt · 发表于 2022-3-20 12:38:58

# 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[int(i)] += 1
result = [i[0] for i in sorted(d.items(), reverse=True, key=lambda d: d[1])][:10]
for i in result:
print(i)

复制代码

isdkz · 发表于 2022-3-20 12:41:11

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[temp] = 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}')
复制代码

yangbaowen · 发表于 2022-3-20 13:55:32

isdkz 发表于 2022-3-20 12:41

大佬

yangbaowen · 发表于 2022-3-20 14:01:01

ckblt 发表于 2022-3-20 12:38

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