解题:-)我有这个名单,想要得到预期的结果,但没有运气。在python和FME中尝试过
如果当前数字与前一个数字之间有2以上,则必须启动一个新行。
有人能帮忙吗?在FME中更好,但在python中也不错
StreetName;HouseNumber;OddEven
Northroad;1;O
Northroad;2;E
Northroad;3;O
Northroad;4;E
Northroad;8;E
Northroad;10;E
Southroad;1;O
Southroad;2;E
Southroad;3;O
Southroad;4;E
Southroad;10;E
Southroad;12;E
Southroad;14;E
我的结果是:
Northroad 1-3
Northroad 2-10
Southroad 1-3
Southroad 2-14
预期结果:
Northroad 1-3
Northroad 2-4
Northroad 8-10
Southroad 1-3
Southroad 2-4
Southroad 10-14
这有什么意义吗?
发布于 2020-04-16 17:14:20
这是一个中等复杂的解析、处理和格式化任务,因此我将使用类:一个用于道路一侧的连续地址范围,另一个用于该范围的容器。这样,加载接口就像添加解码行一样简单,而格式化接口也将一样简单。
代码可以是:
# set the filenames here:
filename = "input.csv"
out = output.txt
class AddressRange:
def __init__(self, row):
self.road = row[0]
self.low = self.high = int(row[1])
self.side = row[2]
def add(self, row):
if (self.road != row[0]) or (self.side != row[2]):
return False
val = int(row[1])
if val < self.low <= val+2:
self.low = val
elif val-2 <= self.high < val:
self.high = val
else:
return False
return True
def __str__(self):
return '%s %d-%d' % (self.road, self.low, self.high)
class AddressList:
def __init__(self):
self.ranges = []
def add(self, row):
for ar in self.ranges:
if ar.add(row):
break
else:
self.ranges.append(AddressRange(row))
def print(self, fd=sys.stdout):
for ar in self.ranges:
print(str(ar), file=fd)
with open(filename) as fd:
rd = csv.reader(fd, delimiter = ';')
lst = AddressList()
_ = next(fd) # skip header line
for row in rd:
if len(row) == 3:
lst.add(row)
with open(out, 'w') as fd:
lst.print(out)
对于示例输入文件,输出如预期的那样:
Northroad 1-3
Northroad 2-4
Northroad 8-10
Southroad 1-3
Southroad 2-4
Southroad 10-14
https://stackoverflow.com/questions/61254436
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