使用 CSV 数组作为 Python 的输入

Question

我收到了一个 csv 文件，其中包含 100 多个数组，我需要运行这些数组来运行我的数据分析代码，但我不知道如何在 Python 中读取这些数组。每个数组前面都有一行，该行仅包含一个整数，该整数给出数组中的行数，并以行“1234567890”结尾，用作行分隔符。

这是 csv 文件的片段：

7,,,,,,,  
1,-199.117,-105.4,-4.525,227.5415,225.2925647,-0.0198891,-2.6547518
2,133.0423,55.4573,-48.4174,155.16,144.1380093,-0.322813,0.3949385
3,129.8405,-16.9527,-303.3192,331.0847,130.9425427,-1.5644458,-0.1298311
4,-73.6373,71.4677,151.517,183.9712,102.616198,1.1678785,2.3711453
5,41.2654,10.4196,30.3773,54.0915,42.5605604,0.6351541,0.2473322
6,-20.3159,-32.4484,62.4574,74.8581,38.2836056,1.2022641,-2.1301853
7,-13.2904,22.029,-28.2895,38.5096,25.7276422,-0.9386666,2.1136489  
1234567890,,,,,,,  
5,,,,,,,  
1,-136.0755,-204.2787,-48.2127,259.2592,245.4512762,-0.1881526,-2.158425
2,220.5184,46.9388,-113.6448,265.1745,225.4586784,-0.4581388,0.2097266
3,-45.3132,169.6283,-49.2729,188.9506,175.576326,-0.2669358,1.8318334
4,-40.7141,34.7414,25.5414,60.9535,53.5219844,0.4465159,2.4351851
5,15.3863,-49.6703,17.1692,56.7635,51.9988166,0.312235,-1.2704018  
1234567890,,,,,,,  
6,,,,,,,   
1,-19.3083,295.4128,191.8666,360.3712,296.0431267,0.5935079,1.6360639
2,-169.8708,-128.3904,-1.0052,215.4187,212.9323449,-0.0046663,-2.4943822
3,15.4505,-209.6656,-178.0715,279.4077,210.2341118,-0.7536439,-1.4972381
4,172.4142,13.0485,-63.7912,192.2842,172.9072576,-0.3447988,0.0755371
5,16.7456,24.8768,-46.5025,55.9188,29.9878358,-1.1933262,0.9783247
6,-8.911,4.1138,12.7751,17.7283,9.8147477,0.9089022,2.7090895  
1234567890,,,,,,,

我确信如果 csv 只是一个大数组，我可以导入该数组，但当我从众多数组中挑选一个数组时，我感到很困惑。在将临时数组替换为 csv 文件中的下一个数组之前，需要对临时数组运行数据分析。

Answer 1

您可以使用 itertools.groupby 将行解析为单独的数组：

import csv
import itertools

with open('errors','w') as err: pass
with open('data','r') as f:
    for key, group in itertools.groupby(
            csv.reader(f),
            lambda row: row[0].startswith('1234567890')):
        if key: continue  # key is True means we've reach the end of an array
        group=list(group) # group is an iterator; we turn it into a list
        array=group[1:]   # everything but the first row is data
        arr_length=int(group[0][0]) # first row contains the length
        if arr_length != len(array): # sanity check
            with open('errors','a') as err:
                err.write('''\
Data file claims arr_length = {l}
{a}
{h}
'''.format(l=arr_length,a=str(list(array)),h='-'*80))
        print(array)

< code>itertools.groupby 返回一个迭代器。它循环遍历 csv.reader(f) 中的行，并将 lambda 函数应用于每一行。当行以 '1234567890' 开头时，lambda 函数返回 True。返回值（例如True 或False）称为key。重要的一点是，itertools.groupby 将返回相同键的所有连续行收集在一起。

Answer 2

这将为您提供一个格式良好的变量，称为“数据”，供您使用。

import csv
rows = csv.reader(open('your_file.csv'))

data = []
temp = []

for row in rows:
    if '1234567890' in row:
        data.append(temp)
        temp = []
        continue
    else:
        temp.append(row)

if temp != []:
    data.append(temp)