如何在具有匹配字符串的嵌套列表中查找索引的最小值和最大值？

Question

我有一个包含潮汐数据的文本文件，前几行是：

Thursday 4 January,11.58,0.38 meters

Thursday 4 January,16.95,0.73 meters

Thursday 4 January,23.68,0.02 meters

Friday 5 January,6.48,0.83 meters

Friday 5 January,12.42,0.33 meters

等，并继续5天以上的数据。你知道吗

我已将此数据转换为嵌套列表，因此它现在打印为：

[['Thursday 4 January', 11.58, 0.38], ['Thursday 4 January', 16.95, 0.73], ['Thursday 4 January', 23.68, 0.02], ['Friday 5 January', 6.48, 0.83], ['Friday 5 January', 12.42, 0.33].....]

以此类推，对于文件中的每一行数据，每个列表的第一个索引是字符串中的日期，第二个和第三个索引是浮动的。你知道吗

对于每个具有匹配的[0]索引（当天）的嵌套列表，我需要找到每个匹配的嵌套列表[2]索引（潮位）的最低和最高浮动，并将它们打印到每天的屏幕上。你知道吗

例如：

Thursday 4 January: 0.02 meters at lowest and 0.73 meters at highest
Friday 5 January: 0.33 meters at lowest and 0.83 meters at highest

因为列表已经按天排序了，所以我想只计算匹配的名称，然后手动进行比较，例如，对于“1月4日星期四”，因为有3个，所以我只比较split\u tides[0-2][2]，找到最小值和最大值，然后每天重复。但我认为一定有一种方法可以自动比较具有匹配字符串的嵌套列表。你知道吗

谢谢你的帮助。你知道吗

编辑：文本文件的图像

Answer 1

首先，您可以使用groupby按日期分组，sort按第三个索引高度分组。此处为示例代码：

    from itertools import groupby
    from operator import itemgetter

    data = [['Thursday 4 January', 11.58, 0.38], ['Thursday 4 January', 16.95, 0.73], ['Thursday 4 January', 23.68, 0.02],
         ['Friday 5 January', 6.48, 0.83], ['Friday 5 January', 12.42, 0.33]]

    for k, g in groupby(data, key=itemgetter(0)):
        a = sorted(g, key=itemgetter(2))
        print('{}: {} meters at lowest and {} meters at highest'.format(k, a[0][2], a[-1][2]))

输出：

Thursday 4 January: 0.02 meters at lowest and 0.73 meters at highest
Friday 5 January: 0.33 meters at lowest and 0.83 meters at highest

顺便说一下，在date中使用groupby之前，请确保您的数据是按日期排序的。你知道吗

希望这能对你有所帮助，如果你有进一步的问题，请发表评论。：）

Answer 2

data = [['Thursday 4 January', 11.58, 0.38], ['Thursday 4 January', 16.95, 0.73], ['Thursday 4 January', 23.68, 0.02], ['Friday 5 January', 6.48, 0.83], ['Friday 5 January', 12.42, 0.33]]

dic={'day':[],'min':[],'max':[]}
for i in data :
    dic['day']+=[i[0]]
    dic['min']+=[i[1]]
    dic['max']+=[i[2]]



from collections import defaultdict
x = defaultdict(dict)

for i in dic['day']:
    x[i]={'min':[],'max':[]}

for i in range(len(dic['min'])):
    x[dic['day'][i]]['min']+=[dic['min'][i]]
    x[dic['day'][i]]['max']+=[dic['max'][i]]

print(dict(x))
"""
structure data for future use

{'Friday 5 January': {'max': [0.83, 0.33], 'min': [6.48, 12.42]},
 'Thursday 4 January': {'max': [0.38, 0.73, 0.02],
  'min': [11.58, 16.95, 23.68]}}
"""



result =[]
for i in x:
    result.append(r'{}: {} meters at lowest an {} meters at highest'.format(i,min(x[i]['min']),max(x[i]['max'])))


print(result)

"""output

['Thursday 4 January: 11.58 meters at lowest an 0.73 meters at highest', 'Friday 5 January: 6.48 meters at lowest an 0.83 meters at highest']
""""

Answer 3

下面是对您有帮助的片段

from pandas import DataFrame

data = [['Thursday 4 January', 11.58, 0.38], ['Thursday 4 January', 16.95, 0.73], ['Thursday 4 January', 23.68, 0.02],
         ['Friday 5 January', 6.48, 0.83], ['Friday 5 January', 12.42, 0.33]]

df = DataFrame.from_records(data)

df.columns = ["Date", "Value", "Height"]

df.groupby(['Date'])['Height'].max()
df.groupby(['Date'])['Height'].min()

输出：

日期 1月5日星期五0.83 1月4日星期四0:73

日期 1月5日星期五0:33 1月4日星期四0:02