richardec

你可以用 data[0::4] 从第一项开始,每四项选择一项, data[1::4] 从第二项开始,每隔第四项进行选择,依此类推。然后,将它们放入数组并使用 dict + zip 使用该数组和列并将其传递给 pd.DataFrame :

df = pd.DataFrame(dict(zip(columns, [data[0::4], data[1::4], data[2::4], data[3::4]])))

输出:

>>> df
      Name  Age     City  Score
0     jack   34   Sydney  155.0
1     Riti   31    Delhi  177.5
2     Aadi   16   Mumbai   81.0
3    Mohit   31    Delhi  167.0
4    Veena   12    Delhi  144.0
5  Shaunak   35   Mumbai  135.0
6    Shaun   35  Colombo  111.0

.str.replace 我能为你做到这一点:

df['Gross'] = df['Gross'].str.replace('^[^\d]+|[^\d]+$', '', regex=True).astype('float')

输出:

>>> df['Gross']
0     0.43  
1     142.50
2     858.37
3     335.45
4     316.83
5     165.36
6     27.33 
7     27.33 
8     426.83
9     53.37 
10    0.35  
11    0.35  
12    108.10
13    515.20
14    390.53
15    390.53
16    543.64
17    159.23
18    159.23
19    159.23
20    12.14 
21    117.62
22    7.74  
23    7.74  
24    56.85 
25    7.00  
26    7.00  
27    96.85 
28    0.40  
Name: Gross, dtype: float64

这应该是有效的:

part1 = df[['COL_1', 'COL_2']].set_axis([0,1,2], axis=1)
part2 = df[['COL_1', 'COL_4']].set_axis([0,1,2], axis=1)
new_df = pd.concat([part1, part2], ignore_index=True)
new_df[3] = new_df[1]
new_df.columns = ['COL_1', 'COL_8', 'COL_9', 'COL_10']

输出:

>>> new_df
  COL_1 COL_8 COL_9 COL_10
0     A  col2  col3   col2
1     B  col2  col3   col2
2     C  col2  col3   col2
3     A  col4  col5   col4
4     B  col4  col5   col4
5     C  col4  col5   col4

我会使用列表理解来拆分每个字符串,然后将其传递给 pd.DataFrame :

lst = ['Fred Green,20/11/2020\n', 'Jack Wilson,01/05/2021\n',]

df = pd.DataFrame([item.strip().split(',') for item in lst], columns=['name', 'date'])

输出:

>>> df
          name        date
0   Fred Green  20/11/2020
1  Jack Wilson  01/05/2021

你先 groupby(['id', 'key']) 和 agg(list) ,可以按索引的第一级进行分组,对于每组,使用 droplevel + to_dict :

new_df = df.groupby(['id', 'key']).agg(list).groupby(level=0).apply(lambda x: x['value'].droplevel(0).to_dict()).reset_index(name='value')

输出:

>>> new_df
   id                              value
0   1  {'a': ['kkk', 'aaa'], 'b': ['5']}
1   2         {'a': ['kkk'], 'b': ['8']}

或者更简单一点:

new_df = df.groupby('id').apply(lambda x: x.groupby('key')['value'].agg(list).to_dict())

pd.Series.map 非常适合这个。

df['AC'].astype(str).map(dict_wu)

结合其他代码:

df['test'] = np.where(df['AC'].astype(str).map(dict_wu) == df['Units'], True, False)

输出:

>>> df
      AC Units   test
2   1002    UG  False
3   1004    MG  False
6   1004    UG   True
7   1005    UG  False
91  1028    MG   True
92  1028    UG  False
93  1028    UG  False

使用如何 x if cond else y 三元语法?

def main_fn(arr_1):
    all_result_summary = []
    for method in ["met_1", "met2", "met_3"]:
        results: ndarray = np.array(main_fn(list(arr_1), method))
        all_result_summary.append(
            pd.DataFrame(
                {
                    "Method": method,
                    "result": results.mean(),
                    "result_sd_ax_0": results.std(ddof=1) if method == "met_3" else results.mean(axis=0).std(ddof=1),
                    "result_sd_ax_1": "NA" if method == "met_3" else results.mean(axis=1).std(ddof=1),
                },
                index=[0],
            )
        )
        summary = pd.concat(all_result_summary, axis=0, ignore_index=True)
    return summary

我不确定是什么 are.equal_to 是的,但这应该对你有用:

fastball = march[march["TaggedPitchType"] == "Fastball"]

这应该是有效的:

df['col'] = df['col'].str.rsplit('.', n=1).str[0].str.split('.').str[-1]

输出:

>>> df
           col
0  common_name
1       amazon
2       amazon
3       amazon
4      walmart
5      walmart
6     michigan

矢量化(读作“更快”)解决方案:

a = np.array(dfa['A'].str.split('').str[1:-1].tolist())
b = np.array(dfb['B'].str.split('').str[1:-1].tolist())

dfb[['disB_1', 'disB_2', 'disB_3']] = (a != b[:, None]).sum(axis=2)

输出:

>>> dfb
    B  disB_1  disB_2  disB_3
0  AC       1       2       1
1  BC       2       1       1
2  CC       2       2       0

用口述把它传给 pd.DataFrame :

dct = {
    'key': pd.Series(key),
    'index0': pd.Series(index0),
    'index1': pd.Series(index1),
    'index2': pd.Series(index2),
}

df = pd.DataFrame(dct)

输出:

>>> df
      key  index0  index1  index2
0  1234.0       1       4       9
1  2345.0       4       3       4
2  2223.0       6       2       6
3  6578.0       3       1       4
4  9976.0       4       6       3
5     NaN       5       8       2
6     NaN       6       5       1
7     NaN       2       3       4
8     NaN       1       1       1

试试这个:

new_df = df.drop(['Level'], axis=1).pivot(columns='Code', index='ID').fillna('').droplevel(0, axis=1).rename_axis(None, axis=1)

输出:

>>> new_df
     Q1    Q2     Q3    Q4      Q5        Q6
ID                                          
1   0.0                                     
2        11.0                               
3              100.0                        
4                     50.0                  
5                           1000.0          
6                                   122000.0