Corralien

按年份和流域名称分组,而不是仅按流域名称分组?

out = rainfall_watershed.groupby(['year', 'WATERSHED_NAME', as_index=False)['inches'].sum()

使现代化 后编辑。

使用 os.environ 得到 SUDO_USER 变量或作为回退 USERNAME .

剧本 test.py :

import os

username = os.environ.get('SUDO_USER', os.environ.get('USERNAME'))
print(os.path.expanduser(f'~{username}'))

执行:

$ python test.py
/home/louis

$ sudo python test.py
/home/louis

从…起 man sudo :

SUDO_USER设置为调用SUDO的用户的登录名。

老生常谈 在编辑之前。

使用 os.path.expanduser :)

>>> os.path.expanduser('~louis')
/home/louis

你可以用 str.replace :

df['GEO'] = df['GEO'].str.replace('POINT\s+\((.*)\s+(.*)\)', r'(\2, \1)', regex=True)
print(df)

# Output
          NAME                                        GEO
0        Paris                           (48.85, 2.31647)
1     New York        (40.731499671618, -73.993457389558)
2          Rio                             (-22.9, -43.2)
3  Airport GRU  (-23.429382262746415, -46.47313507388693)
4         ORLY                   GEOMETRYCOLLECTION EMPTY

你没有使用正确的选项。

从文件中:

陈列max_colwidth设置列的最大宽度。此长度或更长的单元格将被省略号截断。

陈列max_colwidth:int或None 报表中一列的最大字符宽度 数据结构。当列溢出时,会出现“…” 占位符嵌入到输出中。“无”值表示无限制。 [默认值:50][当前值:50]

pd.set_option('display.max_colwidth', None)
print(df)

# Output
0  Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua

尝试:

def sample(df1, df2, num):
    df1_list = [] 
    df2_list = []
    for x in range(1, num+1):
        df1_list.append(df1.sample(n=num))
        df2_list.append(df2.sample(n=num))
    combined_list = df1_list + df2_list
    Baby = pd.concat(combined_list)
    return Baby

# I don't use copy() here because I don't modify E_df or M_df
baby = sample(E_df, M_df, 50)

创建一个 iterator 在你的绳子上:

message = '1:4:a:5:6:7:2:10:72:75:63:6f:6e:74:72:6f:6c:6c:65:72:2e:6f:72:67'

code = iter(message.split(':'))
data = {}

for t in code:
    l = int(next(code), 16)
    d = [next(code) for _ in range(l)]
    data[t] = d

输出:

>>> data
{'1': ['a', '5', '6', '7'],
 '2': ['72', '75', '63', '6f', '6e', '74', '72', '6f', '6c', '6c', '65', '72', '2e', '6f', '72', '67']}

也许你应该放弃 get_queryset 方法A. queryset 这不是一个 context 格言:

class ProductsView(ListView):

    paginate_by = 20
    context_object_name = 'products'
    template_name = "urunler.html"

    def get_queryset(self):
        queryset = super().get_queryset()

        # do stuff here to change queryset dynamically
        ...

        return queryset

    def get_context_data(self, **kwargs):
        context = super().get_context_data(**kwargs)

        # do stuff here to update your context dict
        ...

        return context

尝试 str.findall :

df['0s'] = df['Sequence'].str.findall('^0*').str[0].str.len()
print(df)

# Output:
    Sequence  0s
0  012312312   1
1  024624624   1
2  036901357   1
3  002486248   2
4  045074305   1
5  080666140   1

使用布尔掩码并求和:

N = sum((df['Participant'] == 1) & (df['Condition'] == 1) & (df['RT'].notna()))
print(N)

# Output
1

细节:

m1 = df['Participant'] == 1
m2 = df['Condition'] == 1
m3 = df['RT'].notna()
df[['m1', 'm2', 'm3']] = pd.concat([m1, m2, m3], axis=1)
print(df)

# Output
   Participant  Condition    RT     m1     m2     m3
0            1          1  0.10   True   True   True  # All True, N = 1
1            1          1   NaN   True   True  False
2            1          2  0.48   True  False   True
3            2          1  1.20  False   True   True
4            2          2   NaN  False  False  False
5            2          2  0.58  False  False   True

使用 pd.numeric 作为布尔掩码:

df = df.loc[pd.to_numeric(df['age'], errors='coerce').notna()]
print(df)

# Output
  age
0  55
1  45
2  58

所有非数值都将转换为 NaN .

使用 explode :

cols = ['Summoner', 'Traits', 'Record_ID']
out = df.assign(Record_ID='id_' + df['Rank'].astype(str))[cols] \
        .explode('Traits', ignore_index=True) \
        .rename(columns={'Traits': 'Trait'})
print(out)

# Output:
   Summoner          Trait Record_ID
0     name1    7 Innovator      id_1
1     name1  1 Transformer      id_1
2     name1    3 Enchanter      id_1
3     name1    2 Socialite      id_1
4     name1    2 Clockwork      id_1
5     name1      2 Scholar      id_1
6     name1        2 Scrap      id_1
7     name2       1 Cuddly      id_2
8     name2      1 Glutton      id_2
9     name2    5 Mercenary      id_2
10    name2      4 Bruiser      id_2
11    name2     6 Chemtech      id_2
12    name2      2 Scholar      id_2
13    name2    1 Socialite      id_2
14    name2     2 Twinshot      id_2

这是你所期待的吗?

import matplotlib.pyplot as plt

fig, axs = plt.subplots(5, 5, sharex=True, sharey=True, figsize=(15, 15))
for ax, (district, sr) in zip(axs.flat, g.groupby('RESPONSIBLE DISTRICT')):
    ax.set_title(district)
    ax.plot(sr.index.get_level_values('YEAR'), sr.values)
fig.tight_layout()

plt.show()

你可以用 np.where :

df['price_diff'] = np.where(df['side'] == 0,
                            df['ref_price'] * df['price'],
                            df['ref_price'] * df['price'] * -1)
print(df)

# Output
   side  ref_price  price  price_diff
0     0        100    110       11000
1     1        110    100      -11000

你只需要创造 raw strings (前缀) 'r' )

rolechange = {r"\\Adv":"Adversary",
              r"\\Sci":"Scientist",
              r"\\Inn":"Innocent",
              r"\\Und":"Undetermined"}

>>> df['Role Type'].replace(rolechange)

0       Scientist
1        Innocent
2    Undetermined
Name: Role Type, dtype: object

可能有点慢,因为你的数据很乱:

• 空行
• 空口述
• 字符串值

尝试:

def f_max(d):
    k = v = ''
    if d:
        d = {k: v for k, v in d.items() if isinstance(v, (int, float))}
        if d:
            k = max(d, key=d.get)
            v = d[k]
    return pd.Series({'Name': k, 'Value': v})

df = pd.concat([df, df['Obj_values'].apply(f_max)], axis=1)

输出:

>>> df
                                          Obj_values  Name Value
0               {'Sony': 25, 'Max': 91, 'James': 55}   Max    91
1                            {'Jack': 12, 'Max': 10}  Jack    12
2                                                 {}            
4                                                               
5                           {'Halk': 21, 'Hall': 15}  Halk    21
6                                           {'H': 1}     H     1
7                 {'Aws': 'k', 'Az': 113, 'Gc': 'b'}    Az   113
8  {'Max': 60, 'HBO': 113, 'Sony': 55, 'WS': 1256...    WS  1256

您可以使用理解来提取列表中每个项目的第一个也是唯一一个元素:

df['Volatility expected'] = [v[0] for v in volatility_list]
print(df)

# Output
    Time Currency       Volatility expected                                 Event Actual Forecast Previous
0  02:00      GBP   Low Volatility Expected  U.K. Construction Output (YoY) (Jan)   9.9%     9.2%     7.4%
1  02:00      GBP   Low Volatility Expected       Construction Output (MoM) (Jan)   1.1%     0.5%     2.0%
2  02:00      GBP  High Volatility Expected                             GDP (MoM)   0.8%     0.2%    -0.2%
3  02:00      GBP  High Volatility Expected                             GDP (YoY)  10.0%     9.3%     6.0%

也许这会有帮助:

rng = np.random.default_rng(2022)
df = pd.DataFrame(rng.integers(0, 10, (5, 4)))
sr = pd.Series(rng.integers(0, 10, (5, )))

>>> df
   0  1  2  3
0  7  2  7  0
1  1  6  9  0
2  0  6  8  7
3  8  1  5  0
4  0  4  8  9

>>> sr
0    3
1    9
2    0
3    2
4    6
dtype: int64

>>> df - sr  # does not work
   0  1  2  3   4
0  4 -7  7 -2 NaN
1 -2 -3  9 -2 NaN
2 -3 -3  8  5 NaN
3  5 -8  5 -2 NaN
4 -3 -5  8  7 NaN

>>> df.sub(sr, axis=0)  # work
   0  1  2  3
0  4 -1  4 -3
1 -8 -3  0 -9
2  0  6  8  7
3  6 -1  3 -2
4 -6 -2  2  3

使用 pivot :

out = df.pivot(index='case', columns='val', values='val')
out.columns = [f'val{i}' for i in range(len(out.columns))]

>>> out
     val0 val1 val2
case               
0       x    y    z
1       x  NaN    z
2     NaN    y  NaN