Python - 確定字符串的頻率和進一步處理

Question

我有一些文本文件包含可變列號，由\t（製表符）分隔。類似這樣的：

value1x1 . . . . . . value1xn 
    .  . . . . . . value2xn 
    .  . . . . . .  . 
valuemx1 . . . . . . valuemxn 

我可以掃描並通過下面的代碼確定值的頻率;這

f2 = open("out_freq.txt", 'w') 
f = open("input_raw",'r') 
whole_content = (f.read()) 
list_content = whole_content.split() 
dict = {} 
for one_word in list_content: 
    dict[one_word] = 0 
for one_word in list_content: 
    dict[one_word] += 1 
a = str(sorted(dict.items(),key=func)) 
f2.write(a) 
f2.close() 

和輸出如下：

這

('26047', 13), ('42810', 13), ('61080', 13), ('106395', 13), ('102395', 13)... 

語法是('value', occurence_number)，它按預期工作。我想實現的是：

1. 要轉換輸出語法如下：('value', occurrence_number, column_number)其中列數是發生在input_raw.txt這個值

2. 要使用同一組值的列數出現的次數來分隔欄並將它們寫入不同的文件

Answer 1

如果我理解你想要的東西類似如下：

import itertools as it 
from collections import Counter 

with open("input_raw",'r') as fin, open("out_freq.txt", 'w') as fout: 
    counts = Counter(it.chain.from_iterable(enumerate(line.split()) 
                for line in fin)) 
    sorted_items = sorted(counts.items(), key=lambda x: x[1], reverse=True) 
    a = ', '.join(str((int(key[1]), val, key[0])) for key, val in sorted_items)) 
    fout.write(a) 

請注意，此代碼使用元組作爲關鍵字來區分相同的值（如果它們出現在不同的列中）。從你的問題不清楚這是否可能以及在這種情況下應該做什麼。

用法示例：

>>> import itertools as it 
>>> from collections import Counter 
>>> def get_sorted_items(fileobj): 
...  counts = Counter(it.chain.from_iterable(enumerate(line.split()) for line in fileobj)) 
...  return sorted(counts.items(), key=lambda x:x[1], reverse=True) 
... 
>>> data = """ 
... 10 11 12 13 14 
... 10 9 7 6 4 
... 9 8 12 13 0 
... 10 21 33 6 1 
... 9 9 7 13 14 
... 1 21 7 13 0 
... """ 
>>> with open('input.txt', 'wt') as fin: #write data to the input file 
...  fin.write(data) 
... 
>>> with open('input.txt', 'rt') as fin: 
...  print ', '.join(str((int(key[1]), val, key[0])) for key, val in get_sorted_items(fin)) 
... 
(13, 4, 3), (10, 3, 0), (7, 3, 2), (14, 2, 4), (6, 2, 3), (9, 2, 0), (0, 2, 4), (9, 2, 1), (21, 2, 1), (12, 2, 2), (8, 1, 1), (1, 1, 4), (1, 1, 0), (33, 1, 2), (4, 1, 4), (11, 1, 1)