Remove deprecated compare_dictionaries

qcodes/utils/helpers.py

@@ -41,80 +41,3 @@ def warn_units(class_name: str, instance: object) -> None:
         f"`units` is deprecated for the `{class_name}` "
         f"class, use `unit` instead. {instance!r}"
     )
-
-
-@deprecate("Internal function no longer part of the public qcodes api")
-def compare_dictionaries(
-    dict_1: dict[Hashable, Any],
-    dict_2: dict[Hashable, Any],
-    dict_1_name: Optional[str] = "d1",
-    dict_2_name: Optional[str] = "d2",
-    path: str = "",
-) -> tuple[bool, str]:
-    """
-    Compare two dictionaries recursively to find non matching elements.
-
-    Args:
-        dict_1: First dictionary to compare.
-        dict_2: Second dictionary to compare.
-        dict_1_name: Optional name of the first dictionary used in the
-                     differences string.
-        dict_2_name: Optional name of the second dictionary used in the
-                     differences string.
-    Returns:
-        Tuple: Are the dicts equal and the difference rendered as
-               a string.
-
-    """
-    err = ""
-    key_err = ""
-    value_err = ""
-    old_path = path
-    for k in dict_1.keys():
-        path = old_path + "[%s]" % k
-        if k not in dict_2.keys():
-            key_err += f"Key {dict_1_name}{path} not in {dict_2_name}\n"
-        else:
-            if isinstance(dict_1[k], dict) and isinstance(dict_2[k], dict):
-                err += compare_dictionaries(
-                    dict_1[k], dict_2[k], dict_1_name, dict_2_name, path
-                )[1]
-            else:
-                match = dict_1[k] == dict_2[k]
-
-                # if values are equal-length numpy arrays, the result of
-                # "==" is a bool array, so we need to 'all' it.
-                # In any other case "==" returns a bool
-                # TODO(alexcjohnson): actually, if *one* is a numpy array
-                # and the other is another sequence with the same entries,
-                # this will compare them as equal. Do we want this, or should
-                # we require exact type match?
-                if hasattr(match, "all"):
-                    match = match.all()
-
-                if not match:
-                    value_err += (
-                        'Value of "{}{}" ("{}", type"{}") not same as\n'
-                        '  "{}{}" ("{}", type"{}")\n\n'
-                    ).format(
-                        dict_1_name,
-                        path,
-                        dict_1[k],
-                        type(dict_1[k]),
-                        dict_2_name,
-                        path,
-                        dict_2[k],
-                        type(dict_2[k]),
-                    )
-
-    for k in dict_2.keys():
-        path = old_path + f"[{k}]"
-        if k not in dict_1.keys():
-            key_err += f"Key {dict_2_name}{path} not in {dict_1_name}\n"
-
-    dict_differences = key_err + value_err + err
-    if len(dict_differences) == 0:
-        dicts_equal = True
-    else:
-        dicts_equal = False
-    return dicts_equal, dict_differences