如何统计 文本 中的 空格数 和 数字 的个数 ？

Question

Python新手求教，还望详解！

想要实现实现的功能：
统计一个文本中的空格数和数字的个数。

具体如下：
文本：

4 8 15 16 23 42     520 I LOVE LOST.

得出结果：

number_counts = 7
space_counts = 13  ＃42和520之间有5个空格，4～42之间都是一个1个空格

---

更新部分：
根据 @banagoo 提高的答案，我的代码：（还是有问题？）

import os
os.chdir('/Users/apple/Desktop/Python/chapter')
file_name = "lost.txt"
space_counts = 0
number_counts = 0
with open(file_name, 'r') as f:
      for line in f:
            space_counts += len( line.split() )
            number_list = [x for x in line.split() if x.isdigit()]
            number_counts = len(number_list)

print "space_counts", space_counts
print "number_counts",number_counts

文本：

4 8 15 16 23 42     520 I LOVE LOST.
4 8 15 16 23 42     520 I LOVE YOU.

发现答案不对呀！求指教？

谢谢 @banagoo 的帮助，可运行的代码如下：

import os
os.chdir('/Users/apple/Desktop/Python/chapter')
file_name = "lost.txt"

space_counts = 0
number_counts = 0
number_list = []

with open(file_name, 'r') as f:
    for line in f:
        line = line.strip()
        space_split_list = line.split(' ')
        space_counts += len(space_split_list) - 1
        for word in space_split_list:
                if word.isdigit():
                    number_list.append(word)
        number_counts = len(number_list)

print "space_counts", space_counts
print "number_counts", number_counts

Answer 1

input_str = '4 8 15 16 23 42     520 I LOVE LOST.'
space_split_list = input_str.split(' ')
space_counts = len(space_split_list) - 1
number_list = [x for x in space_split_list if x.isdigit()]
number_counts = len(number_list)

---

import os
os.chdir('/Users/apple/Desktop/Python/chapter')
file_name = "lost.txt"
space_counts = 0
number_counts = 0
with open(file_name, 'r') as f:
    for line in f:
        line = line.strip()  # 每行行首,行尾的空格会被忽略
        space_split_list = line.split(' ')
        space_counts += len(space_split_list) - 1
        number_list = [x for x in space_split_list if x.isdigit()]
        number_counts = len(number_list)

print "space_counts", space_counts
print "number_counts", number_counts

Answer 2

可以尝试用正则表达式。