Question

我有一个名为!input.txt的文件，其中包含多行，每行是0到10之间的随机整数。我想编写一个程序来读取文件并计算每个整数（0-10）在文件中出现的次数。例如，如果我的输入文件中有四个“0”，两个“3”和五个“7”，程序将打印出这样的内容：

Number of occurrences of 0: 4
Number of occurrences of 1: 0
Number of occurrences of 2: 0
Number of occurrences of 3: 2
Number of occurrences of 4: 0
Number of occurrences of 5: 0
Number of occurrences of 6: 0
Number of occurrences of 7: 5
Number of occurrences of 8: 0
Number of occurrences of 9: 0
Number of occurrences of 10: 0

这是我的代码：

mylist = [0,1,2,3,4,5,6,7,8,9,10]
countlist = []
inFile = open("!input.txt", "r")
count = 0

for digit in mylist:
    for line in inFile:
        if digit == int(line):
            count = count + 1
    countlist.append(count)
    count = 0

#Print out the result#
for i in range(11):
    print("Number of occurrences of {0}: {1}".format(i, countlist[i]))

结果如下：

Number of occurrences of 0: 4
Number of occurrences of 1: 0
Number of occurrences of 2: 0
Number of occurrences of 3: 0
Number of occurrences of 4: 0
Number of occurrences of 5: 0
Number of occurrences of 6: 0
Number of occurrences of 7: 0
Number of occurrences of 8: 0
Number of occurrences of 9: 0
Number of occurrences of 10: 0

我想我的嵌套for循环有问题，但我无法弄清楚它是什么。请帮忙。

Answer 1

你的循环的一个问题是你只能在inFile 上循环，因为它是一个打开的文件对象。在第一遍中读完之后，你就在文件的末尾，而且没有更多的行可供阅读。所以变种

inFile = open("input.txt", "r").readlines()
for digit in mylist:
    count = 0
    for line in inFile:
        if digit == int(line):
            count = count + 1
    countlist.append(count)

应该有效。但是，这需要将所有行读入内存，并为每个数字执行循环。更高效的是预先分配空间并进行一次通过：

mylist = [0,1,2,3,4,5,6,7,8,9,10]
countlist = [0]*len(mylist)
inFile = open("input.txt", "r")
for line in inFile:
    digit = int(line)
    countlist[digit] += 1

更好的是，使用defaultdict或Counter：

from collections import Counter
inFile = open("input.txt", "r")
count = Counter()
for line in inFile:
    count[int(line)] += 1

导致半魔法：

with open("input.txt") as fp:
    count = Counter(int(line) for line in fp)

PS：别忘了关闭文件对象。我懒得回去自己做，但是你应该这样做。：^）

Answer 2

以下内容应该有效。关键是使用dict来跟踪数字的出现。

def calc(rst, x):
    rst[int(x.strip())] +=1
    return rst

with open('input.txt','r') as f:
    result = reduce(calc, f, dict(zip(range(11),[0]*11)))

for k,v in result.items(): print k,v

Python嵌套for循环

2 个答案: