小编Dan*_*ejo的帖子

我想要做的是将圆角边缘应用到整个瓷砖，即使是在儿童内部的容器打开时，与折叠时的方式相同。我尝试使用 BoxDecoration 通过其容器应用样式，但它给了我错误。我不知道如何继续，因为与 ListTile 不同的 ExpansionTile 没有形状的属性。

class DocumentTile extends StatelessWidget {
  final Document document;

  const DocumentTile({Key key, this.document}) : super(key: key);
  @override
  Widget build(BuildContext context) {
    return Card(
      margin: const EdgeInsets.only(top: 12, right: 30),
      shape: RoundedRectangleBorder(
        borderRadius: BorderRadius.circular(8.0),
      ),
      color: AppColors.lbBlue.materialColor,
      child: Container(
        width: MediaQuery.of(context).size.width * 0.83,
        child: ExpansionTile(
          tilePadding: const EdgeInsets.only(left: 40.0, right: 30.0),
          backgroundColor: AppColors.nsIconGrey.materialColor,
          trailing: Container(
            width: MediaQuery.of(context).size.width * 0.49,
            child: Row(
              mainAxisAlignment: MainAxisAlignment.spaceBetween,
              children: [
                Row(
                  children: [
                    Container(
                      width: 0.5,
                      height: 50,
                      color: Colors.white, …

我有这个数字列表

list1 = [15,27,48,70,83]

我想要输出

list1 = [12,13,14,15,24,25,26,27,45,46,47,48,67,68,69,70,80,81,82,83]

我知道我可以对每个数字执行此操作，然后将列表合并在一起并对它们进行排序

for i in range(len(list1)):
    list1[i] = list1[i] - 1

有没有更快的方法可以做到这一点？

我有一个txt文档，其结构如下：

1:0.84722,0.52855;0.65268,0.24792;0.66525,0.46562
2:0.84722,0.52855;0.65231,0.24513;0.66482,0.46548
3:0.84722,0.52855;0.65197,0.24387;0.66467,0.46537

第一个带冒号的数字是索引，我不知道打开文件时如何指示它。确实我想把它抹掉。然后数据用逗号和分号分隔，我希望每个数字都在不同的列中，无论分隔符是逗号还是分号。我怎样才能做到呢？

我正在查询公共端点以获取不同交易所的汇率，该汇率返回带有嵌套字典的列表。我最感兴趣的是嵌套字典中的关键“金额”字段。我正在努力想出一个解决方案来存储变量中具有最多“金额”值的嵌套字典。任何想法都会非常有帮助。我正在用头撞墙。

这是列表：

list_with_nested_dicts = [{"partner":"simpleswap","amount":0,"currency":"cel","supportRate":3,"duration":0,"fixed":true,"min":0,"max":0,"exists":false,"id":""},{"partner":"simpleswap","amount":0,"currency":"cel","supportRate":3,"duration":0,"fixed":false,"min":0,"max":0,"exists":false,"id":""},{"partner":"stealthex","amount":37.90346104,"currency":"cel","supportRate":2,"duration":66.62083333333334,"fixed":true,"min":39.91443225,"max":2550.215226,"exists":true,"id":""},{"partner":"stealthex","amount":37.20972396,"currency":"cel","supportRate":2,"duration":23.158333333333335,"fixed":false,"min":77.82938688,"max":25209.49720665,"exists":true,"id":""},{"partner":"godex","amount":0,"currency":"cel","supportRate":0,"duration":0,"fixed":true,"min":0,"max":0,"exists":false,"id":""},{"partner":"changenow","amount":37.2077365,"currency":"cel","supportRate":2,"duration":11.08859649122807,"fixed":false,"min":56.60646516,"max":25259.88795509,"exists":true,"id":""},{"partner":"changelly","amount":0,"currency":"cel","supportRate":1,"duration":0,"fixed":true,"min":0,"max":0,"exists":false,"id":""},{"partner":"changelly","amount":0,"currency":"cel","supportRate":1,"duration":0,"fixed":false,"min":0,"max":0,"exists":false,"id":""},{"partner":"instaswap","amount":0,"currency":"cel","supportRate":2,"duration":0,"fixed":false,"min":0,"max":0,"exists":false,"id":""},{"partner":"exolix","amount":0,"currency":"cel","supportRate":0,"duration":0,"fixed":true,"min":0,"max":0,"exists":false,"id":""},{"partner":"fixedfloat","amount":0,"currency":"cel","supportRate":3,"duration":0,"fixed":false,"min":0,"max":0,"exists":false,"id":""},{"partner":"switchain","amount":0,"currency":"cel","supportRate":2,"duration":0,"fixed":true,"min":0,"max":0,"exists":false,"id":""},{"partner":"changehero","amount":0,"currency":"cel","supportRate":3,"duration":56.41111111111111,"fixed":true,"min":0,"max":0,"exists":false,"id":""},{"partner":"changehero","amount":0,"currency":"cel","supportRate":3,"duration":2.4833333333333334,"fixed":false,"min":0,"max":0,"exists":false,"id":""},{"partner":"binance","amount":0,"currency":"cel","supportRate":2,"duration":0,"fixed":false,"min":0,"max":0,"exists":false,"id":""},{"partner":"nexchange","amount":0,"currency":"cel","supportRate":3,"duration":0,"fixed":true,"min":0,"max":0,"exists":false,"id":""},{"partner":"letsexchange","amount":0,"currency":"cel","supportRate":2,"duration":0,"fixed":true,"min":0,"max":0,"exists":false,"id":""},{"partner":"letsexchange","amount":0,"currency":"cel","supportRate":2,"duration":0,"fixed":false,"min":0,"max":0,"exists":false,"id":""},{"partner":"alfacash","amount":0,"currency":"cel","supportRate":3,"duration":0,"fixed":true,"min":0,"max":0,"exists":false,"id":""}]

数量最多的词典是：

{"partner":"stealthex","amount":37.90346104,"currency":"cel","supportRate":2,"duration":66.62083333333334,"fixed":true,"min":39.91443225,"max":2550.215226,"exists":true,"id":""}

我有一个这样的列表:

A = [[1,2,3,4],[1,1,2,4],[1,2,3,False],[1,False,2,3],[1,2,3,4],[1,2,3,'word'],[5,6,7,8],[1,4,3,4],[True,1,2,4],[0,1,0,1],[0,0,0,0],[False,False,False,False]]

我希望输出像这样的列表:

A = [[1,2,3,4],[1,1,2,4],[1,2,3,4],[5,6,7,8],[1,4,3,4],[0,1,0,1],[0,0,0,0]]

我只想删除或删除任何列表.它有字符串或逻辑的成员.我怎么做.

例如，如果我有字符串“ 4K892”，并将其放入每个字符分开的列表中。如何使用此顺序“ 23456789TJQKA”打印最高字符

伪代码：

list = [4, K, 8, 9, 2]
Highestcharacter(list) = 'K'

我有以下DataFrame作为玩具示例：

a = [5,2,6,8]
b = [2,10,19,16]
c = [3,8,15,17]
d = [3,8,12,20]
df  = pd.DataFrame([a,b,c,d], columns = ['a','b','c','d'])
df

我想创建一个df1仅保留对角元素并将新的上下三角值转换为零的新DataFrame 。

我的最终数据集应如下所示：

    a   b   c   d
0   5   0   0   0
1   0   10  0   0
2   0   0   15  0
3   0   0   0   20

假设我们有一个排序数组，

A = [1,2,2,3,7,7,7,9]

我们希望输出如下所示：

[1]

[2,2]

[3]

[7,7,7]

[9]

这是我的尝试：

def func(A):

    j = 0
    for i in range(len(A)):
        result = []
        while A[i] == A[j] and j < len(A)-1:
            result.append(A[j])
            j += 1
        if result != []:
            print(result)

该函数不包括列表中的最后一个元素，并且运行时间为 O(N^2)，我正在尝试改进。任何帮助将不胜感激。

我四处寻找一个简单的解决方案，但找不到任何有效的方法。我正在尝试减少输入表单中这些 React SVG 图标的线宽，并使图标更薄。这可能吗？

我尝试使用 CSS 笔画和笔划宽度属性，但它不起作用。

我希望将“nums”数组的最后 k 个元素带到第一个元素。比如，输入：

nums = [1,2,3,4,5,6,7], k = 3

输出：

[5,6,7,1,2,3,4]

我有以下代码：

class Solution(object):
    def rotate(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: None Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        k = k % n
        nums[:] = nums[n-k:] + nums[:n-k]

这工作得很好，即将最后 k 个元素带到开头，并且 nums 显示 [5,6,7,1,2,3,4]。但是，一旦我输入以下内容nums = nums[n-k:] + nums[:n-k]，它就会显示生成的 nums 数组与原始数组 [1,2,3,4,5,6,7] 相同。

我的问题是，为什么产量会发生变化？在谷歌搜索和重新编辑该论坛中与“列表引用和复制”相关的某些其他线程时，我可以意识到这与nums = 列表引用有关，但nums[:]就像只是列表的副本。但话说回来，为什么产量会发生这种变化呢？这两个命令内部发生了什么？

我似乎nums还nums[:]不清楚。请帮忙。

我说过三个列表，其中一个是这样嵌套的：

a = [1, 2, 3, 4]
b = ["abc", "def", "dec", "erf"]
c = [[5, 6, 7], [8, 9, 10], [11, 12, 13.3], [14, 15, 16]]

我想要CSV它的文件输出，如下所示：

1,abc,5,6,7
2,def,8,9,10
3,erf,11,12,13.3
...

我尝试将它们压缩并写入CSV文件，如下所示：

1,abc,5,6,7
2,def,8,9,10
3,erf,11,12,13.3
...

但输出有这些愚蠢的括号，如下所示：

1,abc,"[5,6,7]"
2,def,"[8,9,10]"
3,erf,"[11,12,13.3]"
...

但我希望它们没有括号和引号，如下所示：

1,abc,5,6,7
2,def,8,9,10
3,erf,11,12,13.3
...

:(

我正在尝试从数据框中删除值,这是一些值为10.0,10.5,40.0的温度,但是我的值没有意义,如140.0,159.5 ..我想删除它.我使用以下函数,但我没有像索引那样删除任何东西

def remove_outlier(df, col_name):
    threshold = 100.0  # Anything that occurs abovethan this will be removed.
    value_counts = df.stack().value_counts()  # Entire DataFrame
    to_remove = value_counts[value_counts >= threshold].index
    if(len(to_remove) > 0):
        df[col_name].replace(to_remove, np.nan)
    return df

我有三个列表：

gene = [gene_1, gene_2] 
number = [1, 2] 
list_third = ['atcatcg', 'atcatcg'] 

我想创建一个字典，我希望这个字典的键是包含我的第一个列表和第二个列表（基因和数字）的元素的元组，值将是序列

我希望我的输出是这样的：

dict = {(gene_1, 1):'atcatcg', (gene_2, 2):'atcatcg'}