Question

我已拆分字符串以产生以下内容：

Transaction = ['03/29/18', '16,545,275.32-', '8,309,931.13-', '24,855,206.45-', '2.18', '1,505.11']

如何检查列表中的每个字符串以确定它是否以“ - ”结尾，并且是否将其移动到字符串的前面？输出应该类似于：

Transaction = ['03/29/18', '-16,545,275.32', '-8,309,931.13', '-24,855,206.45', '2.18', '1,505.11']

Answer 1

这是一个包含列表理解的单行程序：

>>> Transaction = ['03/29/18', '16,545,275.32-', '8,309,931.13-', '24,855,206.45-', '2.18', '1,505.11']

>>> ['-' + i[:-1] if i[-1] is '-' else i for i in Transaction]
['03/29/18', '-16,545,275.32', '-8,309,931.13', '-24,855,206.45', '2.18', '1,505.11']

说明：

这里i是迭代器，它迭代列表Transaction

中的每个元素

i[-1] - 获取字符串的最后一个字符（适用于非空字符串）

i[:-1] - 从起始元素到最后一个字符的切片字符串（不包括最后一个字符）

与i[-1] is '-'

相比，

python更快地评估表达式i.endswith('-')

In [2]: %timeit ['-' + i[:-1] if i[-1] is '-' else i for i in Transaction]
950 ns ± 7.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [3]: %timeit ['-' + i[:-1] if i.endswith('-') else i for i in Transaction]
1.52 µs ± 10.4 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [4]: %timeit ['-' + i[:-1] if i[-1] == '-' else i for i in Transaction]
1.1 µs ± 4.79 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

如果列表中字符串的最后一个字符是“ - ”，则将其移动到字符串

1 个答案: