The itertools.product() function is for exactly this situation. It takes any number of iterables as arguments and returns an iterator over tuples in the Cartesian product:Python it.product([1, 2], ['a', 'b']) # (1, 'a'), (1, 'b'), (2, 'a'), (2, 'b') ...
注意 在Python 世界中,遵循 Python 哲学的禅使您的代码“Python 化”Python 官方文档中推荐了许多好的实践,以使您的代码更整洁、可读性更好。阅读 PEP8 指南肯定会帮助你理解为什么推荐一些做法。 编写Pythonic 代码 Python 有一些名为 PEP8 的官方文档,定义了编写 python 代码的最佳实践。这种风格指南随着时间的推...
long_set_union_rest | many_duplicates | 0.958 short_set_union_rest | many_duplicates | 0.969 empty_set_union | many_duplicates | 0.971 set_chain | many_duplicates | 1.128 unique_concatenate | many_duplicates | 2.411 func | x | time empty_set_union | many_small_lists | 1.023 long_set_...
使用product 扁平化多层嵌套循环 使用islice 实现循环内隔行处理 使用takewhile 替代 break 语句 使用生成器编写自己的修饰函数 建议2:按职责拆解循环体内复杂代码块 复杂循环体如何应对新需求 使用生成器函数解耦循环体 总结 使用函数修饰被循环对象本身,可以改善循环体内的代码 itertools 里面有很多工具函数都可以用来改...
20.使用“itertools”模块中的“product”来计算多个列表的笛卡尔积，例如：from itertools import product ...
The object returned from zip must be moved into the enumerate object. As a more specific result, itertools can be mixed and nested.Pipe syntaxWherever it makes sense, I've implemented the "pipe" operator that has become common in similar libraries. When the syntax is available, it is done...
您可以通过使用itertools包中的itertools.chain.from_iterable轻松快速的辗平一个列表。下面是一个简单的例子: a_list = [[1,2],[3,4],[5,6]] print(list(itertools.chain.from_iterable(a_list))) # Output: [1, 2, 3, 4, 5, 6] # or print(list(itertools.chain(*a_list))) # Output: [...
# Streamlined approach (using itertools.filter) def streamlined_approach(list1, list2): return list(filter(lambda x: x in list2, list1)) Performance Optimization: Review the code for optimization techniques, especially in computation-heavy applications ...
product(*df.index.levels[:2]), total=len(list(itertools.product(*df.index.levels[:2]))), desc='Computing dt/deep'): subdf = df.loc[f, m] log_age = np.log10(subdf['star_age']) deriv = np.gradient(log_age, subdf['eep']) subdf.loc[:, 'dt_deep'] = deriv df.dt_deep...
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