Question

我有一个csv，其中包含伦敦地铁站的名称和lat / lng位置信息。它看起来像这样：

Station Lat Lng
Abbey Road  51.53195199 0.003737786
Abbey Wood  51.49078408 0.120286371
Acton   51.51688696 -0.267675543
Acton Central   51.50875781 -0.263415792
Acton Town  51.50307148 -0.280288296

我希望转换此csv以创建这些站的所有可能组合的原始目标矩阵。有270个站，因此有72,900个可能的组合。

最终我希望将此矩阵转换为具有以下格式的csv

O_Station   O_lat   O_lng   D_Station   D_lat   D_lng
Abbey Road  51.53195199 0.003737786 Abbey Wood  51.49078408 0.120286371
Abbey Road  51.53195199 0.003737786 Acton   51.51688696 -0.267675543
Abbey Road  51.53195199 0.003737786 Acton Central   51.50875781 -0.263415792
Abbey Wood  51.49078408 0.120286371 Abbey Road  51.53195199 0.003737786
Abbey Wood  51.49078408 0.120286371 Acton   51.51688696 -0.267675543
Abbey Wood  51.49078408 0.120286371 Acton Central   51.50875781 -0.263415792
Acton   51.51688696 -0.267675543    Abbey Road  51.53195199 0.003737786
Acton   51.51688696 -0.267675543    Abbey Wood  51.49078408 0.120286371
Acton   51.51688696 -0.267675543    Acton Central   51.50875781 -0.263415792

第一步是使用循环将任何站与所有其他可能的站配对。然后我需要删除原点和目的地是同一站的0组合。

我尝试过使用NumPy函数column_stack。然而，这给出了一个奇怪的结果。

import csv
import numpy
from pprint import pprint
numpy.set_printoptions(threshold='nan')

with open('./London stations.csv', 'rU') as csvfile:
    reader = csv.DictReader(csvfile)
    Stations = ['{O_Station}'.format(**row) for row in reader]
print(Stations)
O_D = numpy.column_stack(([Stations],[Stations]))
pprint(O_D)

输出

Stations =

['Abbey Road', 'Abbey Wood', 'Acton', 'Acton Central', 'Acton Town']

O_D =

array([['Abbey Road', 'Abbey Wood', 'Acton', 'Acton Central', 'Acton Town',
        'Abbey Road', 'Abbey Wood', 'Acton', 'Acton Central', 'Acton Town']], 
      dtype='|S13')

我理想地寻找更合适的功能，并且难以在Numpy手册中找到它。

Answer 1

这是一个不完整的答案，但我会跳过numpy并向右前进pandas：

csv_file = '''Station Lat Lng
Abbey Road  51.53195199 0.003737786
Abbey Wood  51.49078408 0.120286371
Acton   51.51688696 -0.267675543
Acton Central   51.50875781 -0.263415792
Acton Town  51.50307148 -0.280288296'''

这很难，因为它不是真正用逗号分隔的，否则我们只能拨打pandas.read_csv()：

names = [' '.join(x.split()[:-2]) for x in stations]
lats = [x.split()[-2] for x in stations]
lons = [x.split()[-1] for x in stations]

stations_dict = {names[i]: (lats[i], lons[i]) for i, _ in enumerate(stations)}

df = pd.DataFrame(stations_dict).T    # Transpose it
df.columns = ['Lat', 'Lng']
df.index.name = 'Station'

所以我们最终得到df.head()屈服：

                       Lat           Lng
Station
Abbey Road     51.53195199   0.003737786
Abbey Wood     51.49078408   0.120286371
Acton          51.51688696  -0.267675543
Acton Central  51.50875781  -0.263415792
Acton Town     51.50307148  -0.280288296

获得排列可能意味着我们不需要将电台作为索引...目前还不确定。希望这有点帮助！

Answer 2

当使用这样的表格数据时，我更喜欢使用pandas。它使您的数据结构控制变得简单。

import pandas as pd

#read in csv
stations = pd.read_csv('london stations.csv', index_col = 0)

#create new dataframe
O_D = pd.DataFrame(columns = ['O_Station','O_lat','O_lng','D_Station','D_lat','D_lng'])

#iterate through the stations

new_index= 0
for o_station in stations.index:
    for d_station in stations.index:
        ls = [o_station,stations.Lat.loc[o_station],stations.Lng.loc[o_station],d_station, stations.Lat.loc[d_station], stations.Lng.loc[d_station]]
        O_D.loc[new_index] = ls
        new_index+=1

#remove double stations
O_D = O_D[O_D.O_Station != O_D.D_Station]

这应该可以帮助您进行数据转换。

Python原点目标矩阵表达式中的CSV操作

2 个答案: