[英]How to numerically sort list of strings containing numbers?
f1 = open("leader")
lines = f1.readlines()
lines.sort(key=int, reverse=True)
f1.close()
print(lines)
with external file values: 使用外部文件值:
345345:player7
435:zo
345:name
23:hello
1231:football
this is to sort them so that the integers are sorted not the names 这是为了对它们进行排序,以便整数排序而不是名称
IIUC: IIUC:
l = ['345345:player7',
'435:zo',
'345:name',
'23:hello',
'1231:football']
sorted(l, key=lambda x: int(x.split(':')[0]))
Output: 输出:
['23:hello', '345:name', '435:zo', '1231:football', '345345:player7']
The sort key should do: "split once, convert to integer". 排序键应该:“拆分一次,转换为整数”。 Not converting to integer fails because then lexicographical compare is used and in that case
"10" < "2"
, not that you want. 不转换为整数失败,因为然后使用词典比较,在那种情况下
"10" < "2"
,而不是你想要的。
l = ['345345:player7',
'435:zo',
'345:name',
'23:hello',
'1231:football']
result = sorted(l, key=lambda x: int(x.split(':',1)[0]))
result: 结果:
['23:hello', '345:name', '435:zo', '1231:football', '345345:player7']
that doesn't handle the tiebreaker where numbers are equal. 这不能处理数字相等的决胜局。 A slightly more complex sort key would be required (but still doable).
需要稍微复杂的排序键(但仍然可行)。 In that case, drop
lambda
and create a real function so you can perform split once & unpack to convert only the first part to integer: 在这种情况下,删除
lambda
并创建一个实际函数,这样你就可以执行split一次&unpack将第一部分转换为整数:
def sortfunc(x):
number,rest = x.split(':',1)
return int(number),rest
result = sorted(l, key=sortfunc)
Try this: (helpful if you are still reading from a file) 试试这个:(如果您还在从文件中读取,则会很有帮助)
with open('leader.txt', mode = 'r') as f1:
data = f1.readlines()
# end with
keys = {}
output = []
for s in data:
num, value = s.split(sep=':')
if keys.get(int(num), False):
keys[int(num)].append(value)
else:
keys[int(num)] = [value]
for num in sorted(keys):
for i in keys[num]:
output.append((num, i))
for num, value in output:
print(f'{num}: {value}')
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