list

List Comprehensions in Python to compute minimum and maximum values of a list

此生再无相见时 提交于 2021-02-20 06:25:26
问题 i have the following code to compute minimum and maximum values of a list in order to save memory efficiency x_min = float('+inf') x_max = float('-inf') for p in points_in_list: x_min = min(x_min, p) x_max = max(x_max, p) where points_in_list is a (large) list of numbers. I wish to know if there is a method to compute with a List Comprehensions the min and max value and saving the memory. 回答1: I'm a big fan of generators and comprehensions, but in this case it seems they are not the right way

How to save a list of dataframes to csv

江枫思渺然 提交于 2021-02-20 03:51:46
问题 I have a list of data frames which I reshuffle and then I want to save the output as a csv. To do this I'm trying to append this list to an empty data frame: l1=[year1, year2,..., year30] shuffle (l1) columns=['year', 'day', 'tmin', 'tmax', 'pcp'] index=np.arange(10957) df2=pd.DataFrame(columns=columns, index=index) l1.append(df2) This result in an empty data frames with a bunch of Nans. I don't necessarily need to append my reshuffled list to a dataframe, I just need to save it as a csv, and

Split a dataframe into a list of nested data frames and matrices

ε祈祈猫儿з 提交于 2021-02-20 03:46:53
问题 I'd like to split the diamonds data frame into a list of 5 dataframe, group by cut . This instruction got me started. https://dplyr.tidyverse.org/reference/group_split.html diamonds_g <- diamonds%>% group_split(cut)%>% setNames(unique(diamonds$cut)) My desired output is a list of 5 nested lists. Each nested list contains one data frame and one matrix, such that: View(diamonds_g[[1]]) factors <- diamonds_g[[1]][2:4] mat <- diamonds_g[[1]][6:10] So each of the nested list (or each cut )

Iterate through nested list with many layers

烂漫一生 提交于 2021-02-20 03:43:42
问题 Consider the scenario where you have a collection, and inside that collection are particular objects. Those objects also hold a collection, and inside those collections are more of the same objects. It's a nested collection with many layers. List<WorkItemClassificationNode> items; List<WorkItemClassificationNode> subItems = items.Children; List<WorkItemClassificationNode> subSubItems = subItems.Children; // etc I just want a method to iterate through each of those layers so the same logic is

C# subtracting one list from another or checking if one list is completly containt in another list

烈酒焚心 提交于 2021-02-20 02:26:45
问题 How to subtract one list from another? List<string> l1 = new List<string> { "abc", "abc", "abc", "def" }; List<string> l2 = new List<string> { "abc" }; var r = l1.Except(l2).ToList(); Doing this results in r => "def" instead of r=> "abc", "abc", "def". I mean, the second list only contains "abc" one time. So I want to remove only one instance of "abc" of the first list. Btw: Is there a way to check if one list is completely contained in another list? Meaning when list1 only contains "abc" one

C# subtracting one list from another or checking if one list is completly containt in another list

瘦欲@ 提交于 2021-02-20 02:24:20
问题 How to subtract one list from another? List<string> l1 = new List<string> { "abc", "abc", "abc", "def" }; List<string> l2 = new List<string> { "abc" }; var r = l1.Except(l2).ToList(); Doing this results in r => "def" instead of r=> "abc", "abc", "def". I mean, the second list only contains "abc" one time. So I want to remove only one instance of "abc" of the first list. Btw: Is there a way to check if one list is completely contained in another list? Meaning when list1 only contains "abc" one

C# subtracting one list from another or checking if one list is completly containt in another list

风流意气都作罢 提交于 2021-02-20 02:19:38
问题 How to subtract one list from another? List<string> l1 = new List<string> { "abc", "abc", "abc", "def" }; List<string> l2 = new List<string> { "abc" }; var r = l1.Except(l2).ToList(); Doing this results in r => "def" instead of r=> "abc", "abc", "def". I mean, the second list only contains "abc" one time. So I want to remove only one instance of "abc" of the first list. Btw: Is there a way to check if one list is completely contained in another list? Meaning when list1 only contains "abc" one

C# subtracting one list from another or checking if one list is completly containt in another list

痞子三分冷 提交于 2021-02-20 02:19:26
问题 How to subtract one list from another? List<string> l1 = new List<string> { "abc", "abc", "abc", "def" }; List<string> l2 = new List<string> { "abc" }; var r = l1.Except(l2).ToList(); Doing this results in r => "def" instead of r=> "abc", "abc", "def". I mean, the second list only contains "abc" one time. So I want to remove only one instance of "abc" of the first list. Btw: Is there a way to check if one list is completely contained in another list? Meaning when list1 only contains "abc" one

How to pythonically select a random index from a 2D list such that the corresponding element matches a value?

旧城冷巷雨未停 提交于 2021-02-19 23:44:40
问题 I have a 2D list of booleans. I want to select a random index from the the list where the value is False . For example, given the following list: [[True, False, False], [True, True, True], [False, True, True]] The valid choices would be: [0, 1] , [0, 2] , and [2, 0] . I could keep a list of valid indices and then use random.choice to select from it, but it seems unpythonic to keep a variable and update it every time the underlying list changes for only this one purpose. Bonus points if your

How to pythonically select a random index from a 2D list such that the corresponding element matches a value?

╄→尐↘猪︶ㄣ 提交于 2021-02-19 23:41:08
问题 I have a 2D list of booleans. I want to select a random index from the the list where the value is False . For example, given the following list: [[True, False, False], [True, True, True], [False, True, True]] The valid choices would be: [0, 1] , [0, 2] , and [2, 0] . I could keep a list of valid indices and then use random.choice to select from it, but it seems unpythonic to keep a variable and update it every time the underlying list changes for only this one purpose. Bonus points if your