[英]PYTHON Sort list with empty, string and numeric values
我想對短列表進行排序,例如:
# we can have only 3 types of value: any string numeric value like '555', 'not found' and '' (can have any variation with these options)
row = ['not found', '', '555']
至
# numeric values first, 'not found' less prioritize and '' in the end
['555', 'not found', '']
我嘗試使用
row.sort(key=lambda x: str(x).isnumeric() and not bool(x))
但它不工作
我該如何排序? (數值在前,“未找到”優先級較低,最后是“”)
def custom_sort(list):
L1 = []
L2 = []
L3 = []
for element in list:
if element.isnumeric():
L1.append(element)
if element == 'Not found':
L2.append(element)
else : L3.append(element)
L1.sort()
L1.append(L2).append(L3)
return L1
編輯:按要求對非數值進行排序
ar = [i for i in row if not i.isnumeric()]
ar.sort(reverse=True)
row = [i for i in row if i.isnumeric()] + ar
這將對您的列表進行排序並給予'not found'
比''
更高的優先級:
l = [int(a) for a in row if a.isnumeric()] # Or float(a)
l.sort()
row = [str(a) for a in l] +\
['not found'] * row.count('not found') +\
[''] * row.count('')
這也可以解決問題:
row = ['not found', '', 555, 1, '5' , 444]
print(row)
def func(x):
if str(x).isnumeric():
return 1/-int(x) # ordering numerics event if they are strings
elif str(x) == 'not found':
return 2
elif str(x) == '':
return 3
row2 = row.sort(key=func)
print(row)
結果:
['not found', '', 555, 1, '5', 444]
[1, '5', 444, 555, 'not found', '']
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