基本上我有一个带有数据的csv文件,如下所示:
['Store A', '2015-03-04', '00948', 'Red','A','AA']
['Store C', '2015-05-06', '00948', 'Blue','A','BB']
['Store B', '2015-07-08', '101130', 'Red','B','CC']
['Store A', '2015-09-10', '111011', 'Blue','C','DD']
['Store C', '2015-10-11', '101510', 'Red','A','EE']
['Store B', '2015-11-12', '101459', 'Red','B','FF']
['Store C', '2015-15-04', '01836', 'Blue','C','GG']
['Store B', '2015-30-05', '02201', 'Blue','A','HH']
['Store A', '2015-18-06', '04022', 'Red','C','II']
['Store C', '2015-07-07', '11056', 'Blue','B','JJ']
['Store C', '2015-08-05', '10149', 'Red','D','KK']
['Store A', '2015-10-04', '113569', 'Red','A','LL']
['Store B', '2015-12-03', '005410', 'Blue','C','MM']
['Store A', '2015-15-02', '053410', 'Blue','E','NN']
['Store A', '2015-16-04', '113410', 'Red','J','OO']
我想确定每个列表中出现“蓝色”一词的次数,这样输出基本上是“蓝色”一词的总和,给出第一个属性,即商店A,B和C,需要的输出应该是:
['Store A','Blue','2']
['Store B','Blue','2']
['Store c','Blue','3']
我的代码如下:
csvReader = csv.reader(open('count.csv','rb'), delimiter=',', quotechar='"')
for line in csvReader:
print line.count('Blue')
显然结果是:
>>>
0
0
0
.
.
.
.
0
0
我也试过了代码:
csvReader = csv.reader(open('count.csv','rb'), delimiter=',', quotechar='"')
for line in csvReader:
count_blue= [[x, line.count('Blue')] for x in set(line)]
print count_blue
它也没有给我所需的输出。什么似乎是我的错?谢谢你的帮助。
答案 0 :(得分:1)
这看起来不像是一个CSV文件,它看起来像每行一个Python列表。使用literal_eval
阅读并将其提供给Counter
:
from ast import literal_eval
from collections import Counter
blues = Counter()
with open("count.csv") as f:
for line in f:
ls = literal_eval(line)
if ls[3] == 'Blue':
blues[ls[0]] += 1
如果您想以所需的输出格式打印它:
for key in blues:
print("['{}', 'Blue', {}]".format(key, blues[key]))
答案 1 :(得分:1)
我将假设您的CSV文件实际上是CSV文件。逗号是分隔符,quotechar是单引号char '
。
计算第0列中每个商店发生(从零开始)第3列的次数需要按列0对数据进行分组。一种方法是使用字典。 collections.defaultdict
是一种字典,可以使用公共密钥轻松收集值列表。一旦你有了,你可以产生“蓝色”项目,或“红色”,或任何其他你可能有的计数。
import csv
from collections import defaultdict
d = defaultdict(list)
with open('count.csv') as f:
for row in csv.reader(f, quotechar="'"):
d[row[0]].append(row[3])
for k in sorted(d):
print('{},{}'.format(k, d[k].count('Blue')))
<强>输出强>
Store A,2 Store B,2 Store C,3