如何删除列表中特定元素的最后两个字符？

Question

所以我有这些包含元数据的文本文件，我制作了这段代码来打印它的每一行（第一个元数据文件的）：

path = 'C:\\Users\\basse\\Pictures\\OneM\\metadata\\P5JS metadata\\Metadata'

with open(path + '1.txt') as file:
    lines = []
    for line in file:
        lines.append(line)


print(lines[0], end ="")
print(lines[1], end ="")
print(lines[2], end ="")
print(lines[3], end ="")
print(lines[4], end ="")

现在的问题是第四行和第五行在我的元数据文件中看起来像这样

1.0
4.0

所以当我打印它时会发生什么：

Blue sky
Blue Ice
Rounded Rectangle
1.0
4.0

如何删除这些字符的最后两个字符（So the.0）？

Answer 1

你可以这样做，

with open(path + '1.txt') as file:
    lines = []
    for line in file:
        lines.append(line. replace(".0", ""))

print(lines[0], end ="")
print(lines[1], end ="")
print(lines[2], end ="")
print(lines[3], end ="")
print(lines[4], end ="")

或者你可以这样做

with open(path + '1.txt') as file:
    lines = []
    for line in file:
        lines.append(line)

print(lines[0], end ="")
print(lines[1], end ="")
print(lines[2], end ="")
print(lines[3].replace(".0", ""), end ="")
print(lines[4].replace(".0", ""), end ="")

Answer 2

您可以使用正则表达式替换来去除尾随.0 ：

with open(path + '1.txt') as file:
    lines = []
    for line in file:
        lines.append(re.sub(r'\.\d+$', '', line))

以上实际上将去除任何小数部分。 如果您真的只想定位尾随.0 ，请使用\.0$ 。