python - 如何对带有数字和字母的字符串进行排序以便在python中进行分类？答案

【问题标题】：How to order strings with numbers and letters in order to categorize in python?python - 如何对带有数字和字母的字符串进行排序以便在python中进行分类？
【发布时间】：2020-02-22 02:25:18
【问题描述】：

我正在处理一个数据集，并且我有一些分类在 ICD10 中的诊断。但是，由于我有很多不同的代码，我想将它们分类为更大的类别。所以我在互联网上找到了这些类别。问题是代码就像“A04”或“Z01”，我不能订购它们，因为它们是字母和数字的混合体。我在下面尝试了该代码，但我确实知道变量“diag_icd10_ranges”不正确。有人可以帮帮我吗？

df['code_diag_assoc_icd10'] = df['Assoc_Diagnose']

# Associated category names
diag_icd10_ranges = [(A00, B99), (C00, D49), (D50, D89), (E00, E89), (F01, F99), (G00, G99), 
       (H00, H59), (H60, H95), (I00, I99), (J00, J99), (K00, K95), (L00, L99),
       (M00, M99), (N00, N99), (O00, O9A), (P00, P96), (Q00, Q99), (R00, R99),
       (S00, T88), (V00, Y99), (Z00, Z99)]

diag_icd10_dict = {0: 'infectious_icd10d', 1: 'neoplasms_icd10d', 2: 'blood_icd10d', 3: 'endocrine_icd10d',
           4: 'mental_icd10d', 5: 'nervous_icd10d', 6: 'eye_icd10d', 7: 'ear_icd10d',
           8: 'circulatory_icd10d', 9: 'respiratory_icd10d', 10: 'digestive_icd10d', 11: 'skin_icd10d', 
          12: 'musculo_icd10d', 13: 'genitourinary_icd10d', 14: 'pregnancy_icd10d', 15: 'perinatalperiod_icd10d', 
          16: 'congenital_icd10d',
          17: 'abnormalfindings_icd10d', 18:'injury_icd10d', 19:'morbidity', 20:'healthstatus'}

# Re-code in terms of integer
for num, cat_range in enumerate(diag_icd10_ranges):
df['code_diag_assoc_icd10'] = np.where(df['code_diag_assoc_icd10'].between(cat_range[0],cat_range[1]), 
                                   num, df['code_diag_assoc_icd10'])

# Convert integer to category name using diag_dict
df['cat_diag_assoc_icd10'] = df['code_diag_assoc_icd10'].replace(proc_icd10_dict)

【问题讨论】：

你能举一个你想要的顺序的例子吗？

标签： python dictionary for-loop

【解决方案1】：

您可以使用 bisect_left 并仅使用其下限表示范围：

from bisect import bisect_left

ranges = ["C00","D50","E00","F00","G00","H00","H60","I00",
          "J00","K00","L00","M00","N00","O00","P00","Q00","P00",
          "Q00","R00","S00","V00","Z00"]

def icdGroup(code): return bisect_left(ranges,code)

icdGroup("B20") # 0
icdGroup("H65") # 7

从空白到 = Z00 将位于索引 22。

bisect_left 将为您提供 O(log(22)) 性能，因此如果您有很多代码要分类，这将比顺序搜索更有效。

【讨论】：

【解决方案2】：

你应该能够使用pythonic的方式来做。请参阅下面的代码。

In [21]: diag_icd10_ranges = [{ 1 : ('A00', 'B99') }, 
    ...:                      { 2 : ('C00', 'D49') }, 
    ...:                      { 3 : ('D50', 'D89') }, 
    ...:                      { 4 : ('E00', 'E89') }, 
    ...:                      { 5 : ('F01', 'F99') }, 
    ...:                      { 6 : ('G00', 'G99') }, 
    ...:                      { 7 : ('H00', 'H59') }, 
    ...:                      { 8 : ('H60', 'H95') }, 
    ...:                      { 9 : ('I00', 'I99') }, 
    ...:                      { 10: ('J00', 'J99') }, 
    ...:                      { 11: ('K00', 'K95') }, 
    ...:                      { 12: ('L00', 'L99') }, 
    ...:                      { 13: ('M00', 'M99') }, 
    ...:                      { 14: ('N00', 'N99') }, 
    ...:                      { 15: ('O00', 'O9A') }, 
    ...:                      { 16: ('P00', 'P96') }, 
    ...:                      { 17: ('Q00', 'Q99') }, 
    ...:                      { 18: ('R00', 'R99') }, 
    ...:                      { 19: ('S00', 'T88') }, 
    ...:                      { 20: ('V00', 'Y99') }, 
    ...:                      { 21: ('Z00', 'Z99') }
    ...:                     ]
    ...: 
    ...: heart_failure_icd10_code = 'I50.9'
    ...: 
    ...: chapter_number = [key for rec in diag_icd10_ranges for key, value in rec.items() if value[0] <= heart_failure_icd10_code <= value[1] ]

In [22]: print(chapter_number)
[9]

In [23]:

【讨论】：