Conditions | 11 |
Total Lines | 91 |
Code Lines | 38 |
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Ratio | 0 % |
Changes | 0 |
Small methods make your code easier to understand, in particular if combined with a good name. Besides, if your method is small, finding a good name is usually much easier.
For example, if you find yourself adding comments to a method's body, this is usually a good sign to extract the commented part to a new method, and use the comment as a starting point when coming up with a good name for this new method.
Commonly applied refactorings include:
If many parameters/temporary variables are present:
Complex classes like solph._helpers.create_time_index() often do a lot of different things. To break such a class down, we need to identify a cohesive component within that class. A common approach to find such a component is to look for fields/methods that share the same prefixes, or suffixes.
Once you have determined the fields that belong together, you can apply the Extract Class refactoring. If the component makes sense as a sub-class, Extract Subclass is also a candidate, and is often faster.
1 | # -*- coding: utf-8 -*- |
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53 | def create_time_index( |
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54 | year: int = None, |
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55 | interval_length: float = None, |
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56 | number_of_intervals: int = None, |
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57 | start: datetime.date = None, |
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58 | number: int = None, |
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59 | interval: float = None, |
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60 | ): |
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61 | """ |
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62 | Create a datetime index for one year. |
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63 | |||
64 | Notes |
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65 | ----- |
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66 | To create 8760 hourly intervals for a non leap year a datetime index with |
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67 | 8761 time points need to be created. So the number of time steps is always |
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68 | the number of intervals plus one. |
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69 | |||
70 | Parameters |
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71 | ---------- |
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72 | year : int, datetime |
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73 | The year of the index. |
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74 | Used to automatically set start and number for the specific year. |
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75 | interval_length : float |
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76 | The length time interval in hours e.g. 0.5 for 30min or 2 for a |
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77 | two hour interval (default: 1). |
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78 | number_of_intervals : int |
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79 | The number of time intervals. By default number is calculated to create |
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80 | an index of one year. For a shorter or longer period the number of |
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81 | intervals can be set by the user. |
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82 | start : datetime.datetime or datetime.date |
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83 | Optional start time. If start is not set, 00:00 of the first day of |
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84 | the given year is the start time. |
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85 | |||
86 | Examples |
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87 | -------- |
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88 | >>> len(create_time_index(2014)) |
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89 | 8761 |
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90 | >>> len(create_time_index(2012)) # leap year |
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91 | 8785 |
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92 | >>> len(create_time_index(2014, interval=0.5)) |
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93 | 17521 |
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94 | >>> len(create_time_index(2014, interval=0.5, number=10)) |
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95 | 11 |
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96 | >>> len(create_time_index(2014, number=10)) |
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97 | 11 |
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98 | >>> str(create_time_index(2014, interval=0.5, number=10)[-1]) |
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99 | '2014-01-01 05:00:00' |
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100 | >>> str(create_time_index(2014, interval=2, number=10)[-1]) |
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101 | '2014-01-01 20:00:00' |
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102 | """ |
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103 | if number is not None: |
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104 | warnig_text = "The argument 'number_of_intervals' repleaces 'number'." |
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105 | if number_of_intervals is None: |
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106 | warn(warnig_text) |
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107 | number_of_intervals = number |
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108 | else: |
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109 | raise AttributeError(warnig_text + " You cannot set both!") |
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110 | |||
111 | if interval is not None: |
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112 | warnig_text = "The argument 'interval_length' repleaces 'interval'." |
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113 | if interval_length is None: |
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114 | warn(warnig_text) |
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115 | interval_length = interval |
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116 | else: |
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117 | raise AttributeError(warnig_text + " You cannot set both!") |
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118 | if interval_length is None: |
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119 | interval_length = 1 |
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120 | |||
121 | if number_of_intervals is None: |
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122 | if calendar.isleap(year): |
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123 | hours_in_year = 8784 |
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124 | else: |
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125 | hours_in_year = 8760 |
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126 | number_of_intervals = round(hours_in_year / interval_length) |
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127 | if start is not None: |
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128 | if year is not None: |
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129 | raise ValueError( |
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130 | "Arguments 'start' and 'year' are mutually exclusive." |
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131 | ) |
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132 | else: |
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133 | start = f"1/1/{year}" |
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134 | try: |
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135 | time_index = pd.date_range( |
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136 | start, periods=number_of_intervals + 1, freq=f"{interval_length}h" |
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137 | ) |
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138 | except ValueError: |
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139 | # Pandas <2.2 compatibility |
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140 | time_index = pd.date_range( |
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141 | start, periods=number_of_intervals + 1, freq=f"{interval_length}H" |
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142 | ) |
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143 | return time_index |
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144 |