[英]Convert a csv to a dictionary with multiple values?
我有这样的csv文件:
pos,place
6696,266835
6698,266835
938,176299
940,176299
941,176299
947,176299
948,176299
949,176299
950,176299
951,176299
770,272944
2751,190650
2752,190650
2753,190650
我想将其转换为如下字典:
{266835:[6696,6698],176299:[938,940,941,947,948,949,950,951],190650:[2751,2752,2753]}
然后,在值的范围内填写缺失的数字:
{{266835:[6696,6697,6698],176299:[938,939,940,941,942,943,944,945,946947,948,949,950,951],190650:[2751,2752,2753]}
}
现在,我已尝试使用此处建议的解决方案来构建字典,但是它将新值覆盖旧值。
任何帮助都会很棒。
这是我编写的用于转换csv2dict的函数
def csv2dict(filename):
"""
reads in a two column csv file, and the converts it into dictionary
"""
import csv
with open(filename) as f:
f.readline()#ignore first line
reader=csv.reader(f,delimiter=',')
mydict=dict((rows[1],rows[0]) for rows in reader)
return mydict
最简单的方法是将collections.defaultdict()
与列表一起使用:
import csv
from collections import defaultdict
data = defaultdict(list)
with open(inputfilename, 'rb') as infh:
reader = csv.reader(infh)
next(reader, None) # skip the header
for col1, col2 in reader:
data[col2].append(int(col1))
if len(data[col2]) > 1:
data[col2] = range(min(data[col2]), max(data[col2]) + 1)
当您读取数据时,这也可以实时扩展范围。
根据您的尝试-
from collections import default dict
# open archive reader
myFile = open ("myfile.csv","rb")
archive = csv.reader(myFile, delimiter=',')
arch_dict = defaultdict(list)
for rows in archive:
arch_dict[row[1]].append(row[0])
print arch_dict
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