我正在努力编写以下python脚本。
我有一个csv
文件,看起来像这样
"SNo","Title1","Title2"
"A1-A3,A4-A5","A,B","C"
"A6-A7","X","Y"
"A8","Z","D"
输出应该是应生成
的csv文件"SNo","Title1","Title2"
"A1","A,B","C"
"A2","A,B","C"
"A3","A,B","C"
"A4","A,B","C"
"A5","A,B","C"
"A6","X","Y"
"A7","X","Y"
"A8","Z","D"
我正在阅读文件
cols= [0,1,2]
with open('C:\\down\\1\\list.csv','rb') as f:
reader = csv.reader(f)
for row in reader:
content = list(row[i] for i in cols)
numberlist = content[0].replace("A","").split(",")
print numberlist[0],content[1],content[2]
但我正在努力超越这个。 请告诉我一个很好的方法来解决这个问题。
由于
答案 0 :(得分:2)
你可以试试这个:
output_arr = []
for row in reader:
# Get the mapping for A ranges
idx_map = row[0].replace("A","").split(",")
for row_map in idx_map:
# For each pair or #-#'s creating the min and max idxs
mapping = [int(v) for v in row_map.split('-')]
min_map = min(mapping)
max_map = max(mapping)
for idx in range(min_map,max_map+1):
# For each value in range min_map to max_map, set values of row.
output_arr.append(["A%i"%(idx),row[1],row[2]])
>>> import pprint
>>> pprint.pprint(output_arr)
[['A1', 'A,B', 'C'],
['A2', 'A,B', 'C'],
['A3', 'A,B', 'C'],
['A4', 'A,B', 'C'],
['A5', 'A,B', 'C'],
['A6', 'X', 'Y'],
['A7', 'X', 'Y'],
['A8', 'Z', 'D']]
这将处理异常情况,例如:“A3-A1”
答案 1 :(得分:1)
假设data
是包含csv
文件中数据的列表列表:
data = [["A1-A3,A4-A5","A,B","C"],
["A6-A7","X","Y"],
["A8","Z","D"]]
for line in data:
head, tail = line[0], line[1:]
for range_ in head.split(","):
try:
from_, to = range_.split("-")
c, n, m = from_[0], int(from_[1:]), int(to[1:])
for i in range(n, m+1):
print c + str(i), tail
except:
print range_, tail
输出:
A1 ['A,B', 'C']
A2 ['A,B', 'C']
A3 ['A,B', 'C']
A4 ['A,B', 'C']
A5 ['A,B', 'C']
A6 ['X', 'Y']
A7 ['X', 'Y']
A8 ['Z', 'D']
答案 2 :(得分:0)
我想这可以作为一个起点:
with open('list.csv','rb') as f, open('res.csv', 'wb') as f2:
reader = csv.reader(f)
writer = csv.writer(f2)
for row in reader:
for group in row[0].split(','):
limits = group.split('-')
if len(limits) == 2:
id1, id2 = [int(x[1:]) for x in limits]
for num in range(id1, id2+1):
writer.writerow(("A{}".format(num),row[1],row[2]))
else:
writer.writerow((group,row[1],row[2]))
如果你想要发布的确切输出格式,你需要稍微调整一下。