在python

时间:2019-02-07 06:27:27

标签: python pandas list dataframe group-by

我正在使用python 2.7, 我的文件夹中有文件列表,有数千个文件,看起来像这样:

20180828-024308.dat
20180828-024434.dat
20180828-030335.dat
20180828-032114.dat
20180828-040041.dat
..........

它们是年,月,日,时,分和秒

我想将所有这些文件分组为半小时间隔(注意:年,月和日没有变化)

我想要这样的东西:

1: [20180828-024308.dat,20180828-024434.dat]
2: [20180828-030335.dat,20180828-032114.dat]
3: [20180828-040041.dat,....]
.......

我认为列表对我来说很好,或者对数据框来说可能。

谢谢您的帮助!

3 个答案:

答案 0 :(得分:3)

来自:-据我了解, 假设您的数据框看起来像:

Boolean x = true;
int y = 1;
int z = 1;
if(y ==1 && x == true){
    z++;
    x = false;
}
else{
    z--;
    x = true;
}

输出:

print(df)

                 files
0  20180828-024308.dat
1  20180828-024434.dat
2  20180828-030335.dat
3  20180828-032114.dat
4  20180828-040041.dat

df['file_time']= pd.to_datetime(df['files'].str.split('.dat').str[0])
df.groupby([pd.Grouper(key='file_time',freq='1800s')])['files'].apply(list).reset_index()

注意:由于在3:30-4范围内没有文件,因此列表为空。

答案 1 :(得分:0)

我认为您也可以通过基本编程来实现。 因此,首先使用os库加载所有文件,然后使用python获取文件列表。 这是我要说的一句话

import os

folderPath = '/somepath'
filesInFolder = os.listdir(folderPath)
mapOfsimmilarFiles = {}
keyForMaps = 0
for fileNames in sorted(filesInFoldeyr):
    timePartOfFile = fileNames.split('-')[-1].split('.dat')[0]
    hr = timePartOfFile[0:2]
    min = timePartOfFile[2:4]
    sec = timePartOfFile[4:]
    if len(mapOfsimmilarFiles.keys()) == 0:
        mapOfsimmilarFiles[hr+'_'+min] = [fileNames]
    else:
        minsPresentInMaps = mapOfsimmilarFiles.keys()
        hrPresent = [int(h.split('_')[0]) for h in mapOfsimmilarFiles]
        minPresent = [(h.split('_')[1]) for h in mapOfsimmilarFiles]
        for timeUsed in minsPresentInMaps:
            hrPresent = timeUsed.split('_')[0]
            minPresent = timeUsed.split('_')[1] 
            if abs(int(hrPresent)-int(hr)) == 1:
                if abs(int(minPresent)-int(min)) <=30:
                    mapOfsimmilarFiles[timeUsed].append(fileNames)
                else:
                    #same hr but not 30mins so add to map as a new entry
                    mapOfsimmilarFiles[hr+'_'+min] = [fileNames]
                break
        mapOfsimmilarFiles[hr+'_'+min] = [fileNames]            

我希望这会对您有所帮助,并指导您朝正确的方向前进。

答案 2 :(得分:0)

首先将您的数据转换为dict,然后相应地连接这些字符串。

代码:

d = ['20180828-024308.dat', '20180828-024434.dat', '20180828-030335.dat', '20180828-032114.dat', '20180828-040041.dat']

output = {}

for i in d:
    key = i.split('-')[0]
    key1 = i.split('-')[1]
    # print(output)
    if key in output:

        if key1[0:2] in output[key]:

            output[key][key1[0:2]].append(key1[2:])
        else:
            output[key][key1[0:2]] = [key1[2:]]
    else:
        output[key] = {}
        output[key][key1[0:2]] = [key1[2:]]

print(output)
# print("_".join("{}_{}".format(k, v) for k, v in output.items()))
main_output = []
for i in output.keys():
    temp = []
    for j in output[i].keys():
        # [s + mystring for s in mylist]
        temp.append([i + '-' + j + s for s in output[i][j]])
    main_output.extend(temp)

print(main_output)

输出:

{'20180828': {'02': ['4308.dat', '4434.dat'], '03': ['0335.dat', '2114.dat'], '04': ['0041.dat']}}
[['20180828-024308.dat', '20180828-024434.dat'], ['20180828-030335.dat', '20180828-032114.dat'], ['20180828-040041.dat']]