協助計算文件中的數字

Question

我試圖讓alist中的值成爲文件中0-9的出現次數。我不知道我在哪裏搞砸了，因爲在運行代碼之後，alist仍然全是0。

def main(): 
    intro() 
    inFile = getFile() 
    file, outfile = convertName(inFile) 
    alist, count = countLines(file, outfile) 
    printResults(alist, count, outfile) 

def intro(): 
    print() 
    print("Program to count letters in each line") 
    print("in a file.") 
    print("You will enter the name of a file.") 
    print("The program will create an output file.") 
    print("Written by .") 
    print() 
def getFile(): 
    inFile = input("Enter name of input file: ") 
    return inFile 
def convertName(inFile): 
    file = open(inFile, "r") 
    outfile = (inFile.replace(".txt", ".out")) 
    return file, outfile 
def countLines(file, outfile): 
    outfile = open(outfile, "w") 
    alist = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 
    count = 0 
    for line in file: 
     count = count + 1 
     spl = line.split()  
     for i in range(0,10): 
      for ch in spl: 
       if ch == i: 
        alist[i] = alist[i+1] 
    return alist, count 
def printResults(alist, count, outfile): 
    print("The name of output file is", outfile) 
    print() 
    print("Number of lines:   ", count) 
    t = 0 
    print(alist) 
main() 

Answer 1

for i in range(0,10): 
     for ch in spl: 
      if ch.isdigit() and int(ch) == i: 
       alist[i] += 1 

Answer 2

在：

alist[i] = alist[i+1] 

要指定的元素0至alist[i]。您可能需要做一些事情，如：

Answer 3

如果你只是想那些數（不計的數字），使用collections.Counter

from collections import Counter 
def count_digits(file_name): 
    with open(fn): 
     c = Counter([int(character) for character in open(fn).read() if character.isdigit()]) 
    return [c.get(i,0) for i in range(10)] 

，如果你也想的行數，延長方法有點：

from collections import Counter 
def count_digits(file_name): 
    with open(fn) as fh: 
     c = Counter([int(character) for character in fh.read() if character.isdigit()]) 
     digit_counts = [c.get(i,0) for i in range(10)] 
     fh.seek(0) 
     lines_count = sum(1 for line in fh) 
    return digit_counts, line_counts