如何舍入嵌套元组列表中的每个浮点数

I don't know about "quickest" (quickest to write? read? runtime?), but this is how I'd write it recursively:def re_round(li, _prec=5): try: return round(li, _prec) except TypeError: return type(li)(re_round(x, _prec) for x in li)演示:x = [[(-88.99716274669669, 45.13003508233472), (-88.46889143213836, 45.12912220841379), (-88.47075415770517, 44.84090409706577), (-88.75033424251002, 44.84231949526811), (-88.75283245650954, 44.897062864942406), (-88.76794136151051, 44.898020801741716), (-88.77994787408718, 44.93415662283567), (-88.99624763048942, 44.93474749747682), (-88.99716274669669, 45.13003508233472)]]re_round(x)Out[6]: [[(-88.99716, 45.13004), (-88.46889, 45.12912), (-88.47075, 44.8409), (-88.75033, 44.84232), (-88.75283, 44.89706), (-88.76794, 44.89802), (-88.77995, 44.93416), (-88.99625, 44.93475), (-88.99716, 45.13004)]](该函数的旧生成器版本，用于后代:)(old generator version of the function, for posterity:)def re_round(li, _prec=5): for x in li: try: yield round(x, _prec) except TypeError: yield type(x)(re_round(x, _prec)) 这篇关于如何舍入嵌套元组列表中的每个浮点数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！