Question

我正在尝试拆分一些数据，数据采用这种形式......

['20150406,34.4800,34.8100,34.2300,34.4200,21480500', '20150407,34.5400,34.8900,34.5100,34.6300,14331200']

列表中每个字符串中的第一项是日期，我正在尝试在选定日期拆分数据。但是有整个字符串...例如，如果我选择的日期是2015-04-07，那么上面的数据会像这样分开......

['20150406,34.4800,34.8100,34.2300,34.4200,21480500']

['20150407,34.5400,34.8900,34.5100,34.6300,14331200']

这也必须适用于包含大量字符串的列表，其形式与此相同......

Answer 1

使用next()和enumerate()查找包含所需日期的字符串位置，然后 slice ：

next()

情侣笔记：

StopIteration将通过try/except例外 - 您应该使用-1处理它，或者提供默认值next((i for i, item in enumerate(l) if item.startswith(d)), -1)例如：
```
date
```
检查日期是否与所需日期匹配，我们只是检查项目是否以特定日期字符串开头。如果所需日期为datetime或strftime()，则需要事先使用>>> from datetime import datetime >>> d = datetime(2015, 4, 7) >>> d = d.strftime("%Y%m%d") >>> d '20150407'对其进行格式化：
```
rotate90 = (map reverse .) $ foldr (zipWith (:)) $ repeat ""
```

Answer 2

我认为你想要一个groupby，将不以日期开头的字符串分组，并且这样做的日期会分隔这些组：

l = ['20150406,34.4800,34.8100,34.2300,34.4200,21480500', '2015010,34.5400,34.8900,34.5100,34.6300,14331200'
    , '20150407,34.5400,34.8900,34.5100,34.6300,14331200']
dte = "2015-04-07"

delim = dte.replace("-","") + ","
from itertools import groupby

print([list(v) for k,v in groupby(l,key=lambda x: not x.startswith(delim))])

[['20150406,34.4800,34.8100,34.2300,34.4200,21480500', '2015010,34.5400,34.8900,34.5100,34.6300,14331200'],   ['20150407,34.5400,34.8900,34.5100,34.6300,14331200']]

groupby将继续将数据拆分多次，因为字符串以日期开头。

Answer 3

从alecxe回答延伸：代码可以按输入日期将原始列表拆分为耦合子列表。

l = [
...      '20150406,34.4800,34.8100,34.2300,34.4200,21480500',
...      '20160402,34.1,32.8100,33.2300,31.01,22282510',
...      '20150407,34.5400,34.8900,34.5100,34.6300,14331200',
...      '20120101,2.540,14.8201,32.00,30.1311,12331230',
...  '20150407,34.5400,34.8900,34.5100,34.6300,14331200',]

index = [i for i, item in enumerate(l) if item.startswith(d)]
[l[i:j] for i, j in zip([0]+index, index+[None])]

output:
[['20150406,34.4800,34.8100,34.2300,34.4200,21480500', '20160402,34.1,32.8100,33.2300,31.01,22282510'], ['20150407,34.5400,34.8900,34.5100,34.6300,14331200', '20120101,2.540,14.8201,32.00,30.1311,12331230'], ['20150407,34.5400,34.8900,34.5100,34.6300,14331200']]

分裂数据 - 特定情况

3 个答案: