计算数字在文件中出现的次数

Question

So I'm taking a file and running it through my code. Example of what appears on file per line:

100
200
300
100
200
400

My goal is to get my code to iterate through the numbers in the file and the output is a dictionary with the number as the key and how ever many times it appears in the file as the value. For example:

{100:2,200:2,300:1,400:1}

This is what I've put together so far.

def counts(filename):
    d={}
    with open(filename) as f:
        for line in f
            for number in line:


    return d

Also,would I be able to use .count() for this? So could I create a list of numbers in the file and set those as the keys and then have a list for the set the corresponding amount of times each number appears and set that as the values to the keys?

Answer 1

def counts(filename):
    d={}
    with open(filename) as f:
        contents = f.read()

    contents = contents.split("\n")
    del contents[-1]
    contents =  map(int, contents)

    for content in contents:
        if content not in d:
            d[content] = 1
        else:
            d[content] = d[content] + 1

    return d


print counts(filename)

o/p

{200: 2, 300: 1, 400: 1, 100: 2}

Answer 2

u can create a matrix where each position could represent a number and its content represent the number it appears in the file. furthermore, u can create a comparator that compares with the numbers from the file and then increase the counter

Answer 3

For each number in the file, if it does not appear as a key in your dictionary, add it (with a count of 0); regardless, increment the count for that number.

Answer 4

This uses a generator to read all of the lines and converts them to integers.

from collections import Counter
from csv import reader

def counts(filename):
    return Counter(int(line[0]) for line in reader(open(filename)) if line)


c = counts('my_file.csv')
>>> c
Counter({'100': 2, '200': 2, '300': 1, '400': 1})

>>> c.most_commont(5)
[('200', 2), ('100', 2), ('300', 1), ('400', 1)]

>>> dict(c)
{'100': 2, '200': 2, '300': 1, '400': 1}

Answer 5

One simple way is to use a Counter , which is easily convertible to a dict once the counting is done.

from collections import Counter 

def counts(filename):
    with open(filename) as f:
        return dict(Counter(int(line) for line in f))

# {200: 2, 100: 2, 300: 1, 400: 1}

Answer 6

I would use a defaultdict to keep track of your number counts:

from collections import defaultdict
frequencies = defaultdict(int)
for number in open('numbers.txt'):
    frequencies[int(number)] += 1

for number in sorted(frequencies.keys()):
    print(number, ':', frequencies[number])

Gives:

100 : 2
200 : 2
300 : 1
400 : 1

With a regular dictionary, you need to catch the KeyError on the first time the number is encountered:

count = {}
for number in open('numbers.txt'):
    try:
        count[int(number)] += 1
    except KeyError:
        count[int(number)] = 1

Answer 7

Using only simple python:

word_count = {}
with open('temp.txt') as file:
    for line in file:
        word_count[line[:-1]] = word_count.setdefault(line[:-1], 0) + 1

If you wish to use fancy libraries, you can go with @alexander's answer with Counter .