Abhyuday Vaish

不要用Python关键字命名列表。 list 是Python中的一个关键字。你可以使用列表理解来使用 .startswith() :

list1 = ['abc_apple','abc_orange','cherry','abc_berry','banana']
list1 = ['abc_'+i if not i.startswith('abc_') else i for i in list1]
print(list1)

输出:

['abc_apple', 'abc_orange', 'abc_cherry', 'abc_berry', 'abc_banana']

第一圈 Indices 获取行和列索引,即。 i0 和 i1 然后分别获取 A .用这个:

output = np.array([[A[i0][i1]] for i0, i1 in Indices])

输出:

array([[2.3],
   [1.4],
   [2.5]])

您可以将列表转换为 np.array 然后减去 val :

import numpy as np

df['A_new'] = df['A'].apply(lambda x: np.array(x) - val)

输出:

           A      A_new
0     [1, 2]     [0, 1]
1  [4, 5, 6]  [3, 4, 5]

如果您想获得不同的数据帧,那么以下是您可以做到的:

df_splits = [v for k, v in df.groupby('Name')]
for df_split in df_splits:
    print(df_split, sep = '\n')

输出:

  Name  VALUE
0    A      1
1    A      2
2    A      3
  Name  VALUE
3    B      4
4    B      5
  Name  VALUE
5    C      6
6    C      7
7    C      8
8    C      9
  Name  VALUE
9    D     10

要单独访问数据帧,请使用:

df_splits[0] , df_splits[1] ....

以下是如何重命名该列:

mtcars.rename(columns={mtcars.columns[0]: 'car'}, inplace = True)
# This is equivalent to
mtcars = mtcars.rename(columns={mtcars.columns[0]: 'car'})

输出:

                    car   mpg  cyl   disp   hp  drat     wt   qsec  vs   am
0             Mazda RX4  21.0    6  160.0  110  3.90  2.620  16.46   0  1.0
1         Mazda RX4 Wag  21.0    6  160.0  110  3.90  2.875  17.02   0  1.0
2            Datsun 710  22.8    4  108.0   93  3.85  2.320  18.61   1  1.0
3        Hornet 4 Drive  21.4    6  258.0  110  3.08  3.215  19.44   1  0.0
4     Hornet Sportabout  18.7    8  360.0  175  3.15  3.440  17.02   0  0.0
5               Valiant  18.1    6  225.0  105  2.76  3.460  20.22   1  0.0
6            Duster 360  14.3    8  360.0  245  3.21  3.570  15.84   0  0.0

看看你是否会使用 int 然后不能将货币与美元符号连接起来,因为“$”是一个字符串。 你可以这样做:

# Trying to make it have a special character when file is printed such as "$" Example: $2600
savings = '$' + input("How much money do you have in your savings: ")
deposits = '$' + input("How much money do you put in deposits: ")

只需创建一个新的空str变量并连接它。

str5 = 'peter piper picked a peck of pickled peppers.'
b = str5.split()
rev_str5 = ""
for i in b:
    rev_str5 = rev_str5 + ' ' + i[::-1]
print(rev_str5.lstrip()) # Removes the one space in the starting.

这里还有一个简短的方法。感谢您的评论:

str5 = 'peter piper picked a peck of pickled peppers.'    
print(' '.join(w[::-1] for w in str5.split()))

输出:

retep repip dekcip a kcep fo delkcip .sreppep