在Python中计算文本文件中特定字母或符号

Question

我正在尝试让Python计算一个文本文件中某个字母或符号出现的次数。我的文本文件内容是'*#%##'，但不知为什么，当我输入一个符号时，它却把所有字符都算上了，所以如果我输入'#'，输出的结果是5，而不是3。

这是我到目前为止所做的：

Symbol = input("Pick a Symbol ")
freq = 0
with open ("words.txt", "r") as myfile:
    data = myfile.read().replace('\n', '')
    print(data)
    for Symbol in data:
        freq = (freq + 1)
print(freq)

Answer 1

对于一个很大的输入文件，你可能想要考虑使用 collections.Counter 这个工具。

from collections import Counter

def countsymbols(filename,symbols):
    """Counts the symbols in `filename`.
`symbols` is an iterable of symbols"""
    running_count = Counter()

    with open(filename) as f:
        for line in f:
            running_count += Counter(line.strip())

    return {key:value for key,value in running_count.items() if key in symbols}

symbols = map(str.strip,input("Enter symbols: ").split())
filename = input("Filename: ")

symbolcount = countsymbols(filename,symbols)

Answer 2

你在 for 循环里重新定义了 Symbol：

for Symbol in data:

这段代码只是把文件里的每个字符都赋值给 Symbol，然后计数加一。

建议使用 str.count() 方法，这样更简单：

with open ("words.txt", "r") as myfile:
    data = myfile.read().replace('\n', '')
    print(data)
    freq = data.count(Symbol)
    print(freq)

或者，如果你非得用循环的话，那就逐个检查每个字符：

with open ("words.txt", "r") as myfile:
    data = myfile.read().replace('\n', '')
    print(data)
    freq = 0
    for char in data:
        if char == Symbol:
            freq = freq + 1
    print(freq)