Is there a method to make dictionary lexicographical order ignoring upper and lowercase?

lexicographic order python
lexicographical order java
lexicographical order c++
lexicographic order algorithm
lexicographic order of numbers
lexicographical order in sql
lexicographically smallest string in java
lexicographically vs alphabetically

Is there a method to make dictionary lexicographical order ignoring upper and lowercase? Now I know make a dictionary lexicographical order but still confused how to ignore lowercase and uppercase. Here are some codes I have tried.

dic={'subject':'math','name':'','PS':'$'}
for key,value in sorted(dic.items()):
    print(key+'='+value)

Now I have a result that PS=$ is in the first, but I wanna it in the middle, like this:

name=
PS=$
subject=math

In Python 3.6 and 3.7 the dictionaries are insertion ordered, however it is preferable to use OderedDict

Are dictionaries ordered in Python 3.6+?

from collections import OrderedDict

dic={'subject':'math','name':'','PS':'$'}
dic_ord = OrderedDict(sorted(dic.items(), key=lambda t: t[0].lower()))

for key,value in dic_ord.items():
    print(key, '=', value)

To test whether two strings are equal to each other, you must use the method The expression Do not use the == if (string1 == string2) // Not useful operator to compare strings. The compareTo method compares strings in dictionary order. by putting numbers first, then uppercase characters, then lowercase characters​. To put items in order, there must be a way to compare two items. With strings, the usual order is Lexicographic Order. This is dictionary order, except that all the uppercase letters preceed all the lowercase letters. This order is what the compareTo() method of class String uses.

In-place: Impossible

Reason : Once a key is modified, it won't be possible to retrieve back the original value out of the same old dictionary with the modified key. The only way out here is to store the modified key along with original key.

Create a temporary dictonary, say dic_lex.

Store the modified key (lower-cased) as key and original key as value:

dic={'subject':'math','name':'','PS':'$'}
dic_lex={key.lower() : key for key in dic.keys()}
for key in sorted(dic_lex.keys()):
    print(dic_lex[key]+'='+dic[dic_lex[key]])

prints:

name=
PS=$
subject=math

Each of these routines compares the src1 string to the src2 string and returns true to ignore alphabetic case in string comparisons because different users have than they treat uppercase characters the same as their lowercase equivalents, lexicographical ordering corresponds to the way words appear in a dictionary. An Yan. Apparently, this user prefers to keep an air of mystery about them. 2 Is there a method to make dictionary lexicographical order ignoring upper and lowercase?

Use sorted with the dictionary constructor

dic = dict(sorted(dic.items(), key=lambda x: x[0].lower()))
# {'name': '', 'PS': '$', 'subject': 'math'}

For example: In this case, every time Col1 values have to be compared, the SQL way to compare strings and produces a dictionary-like ('lexicographic') ordering. whether to ignore punctuation, whether to fold upper and lower case, how to  In mathematics, the lexicographic or lexicographical order (also known as lexical order, dictionary order, alphabetical order or lexicographic(al) product) is a generalization of the way words are alphabetically ordered based on the alphabetical order of their component letters.

In mathematics, the lexicographic or lexicographical order is a generalization of the way words In dictionary order, the word "Thomas" appears before "​Thompson" because the letter 'a' comes elements which have the same rank in the sequences, the lexicographical order extends to Cartesian products of ordered sets. Largest lexicographical string with at most K consecutive elements; Make the string lexicographically smallest and non palindromic by replacing exactly one character; Generate an array of given size with equal count and sum of odd and even numbers; Find the k-th string in lexicographical order consisting of n-2 X's and 2 Y's

Sorting should have been easy. This is dictionary order, except that all the uppercase letters preceed all the lowercase This order is what the compareTo​() method of class String uses. All uppercase letters come before lower case letters. Given two strings (of lowercase letters), a pattern and a string. The task is to sort string according to the order defined by pattern. It may be assumed that pattern has all characters of the string and all characters in pattern appear only once. Input : pat = "bca", str = "abc" Output : str = "bca" Input : pat = "bxyzca", str = "abcabcabc

In UTF8, ASCII characters have the lowest possible values, so non-ASCII characters all have lower byte values than their lowercase equivalents, so they don't sort together. Basically this just means that lexicographic sorting is not alphabetic sorting, and We'll sort according to the English alphabet and ignore accents. Key words: Decision making, lexicographical procedure, Superiority degree, Preference relations. 1. Introduction The lexicographical procedure widely spread in the decision making problems. It is used in the various sphere of the human action. for example: 1.The words are lexicographically regulated in the dictionary.

Comments
  • Dictionaries are never lexiographically ordered; the order is arbitrary prior to Python 3.6/3.7, then the order is based remembering the insertion order
  • But what you're actually asking it seems is "can I pass a key to sorted() to ignore the case", the answer is yes of course (like sorted(dict.items(), key = lambda x: x[0].casefold())); and you could feed that to an OrderedDict to retain the sorted structure
  • If you're only looking to get the ordering during printing, you could try something like for k, v in sorted(dic.items(), key=lambda x: x[0].lower()): but obviously, anything that's printed it then thrown away, the original dictionary itself will not have a new order
  • @Chris_Rands So the code I write can't make what I print is lexicographical order?
  • You can look into lambda functions. As a super brief explanation, you are saying "I want you to sort by the first item in the tuple - the key - after you cast it to lowercase". The key argument will take lambda functions and apply them to allow you to change the default sorting rule