我已经编写了一些代码,可以根据多种条件模拟熊猫数据框中的值。现在,我只想为df['Use Type']
列中的特定值运行此代码。我目前有以下内容:
def l_sim():
n = 100
for i in range(n)
df['RAND'] = np.random.uniform(0, 1, size=df.index.size)
conditions = [df['RAND'] >= (1 - 0.8062), (df['RAND'] < (1 - 0.8062)) & (df['RAND'] >= 0.1),
(df['RAND'] < 0.1) & (df['RAND'] >= 0.05), (df['RAND'] < 0.05) &
(df['RAND'] >= 0.025), (df['RAND'] < 0.025) & (df['RAND'] >= 0.0125),
(df['RAND'] < 0.0125)]
choices = ['L0', 'L1', 'L2', 'L3', 'L4', 'L5']
df['L'] = np.select(conditions, choices)
conditions = [df['L'] == 'L0', df['L'] == 'L1', df['L'] == 'L2', df['L'] == 'L3',
df['L'] == 'L4', df['L'] == 'L5']
choices = [df['A'] * 0.02, df['A'] * 0.15, df['A'] * 0.20, df['A'] * 0.50,
df['A'] * 1, df['A'] * 1]
df['AL'] = np.select(conditions, choices)
l_sim()
如何仅对具有df.loc[df['Use Type'] == 'Commercial Property']
的行运行此代码?
谢谢。
答案 0 :(得分:1)
我认为您需要以不同的方式构造代码。但通常,您可以使用df.apply
和lambda函数。此模式:
df['L'] = df.apply(lambda row: l_sim(row), axis=1)
我会将您的代码分成三个函数,一个用于df['L']
:
def l_logic():
random_num = np.random.uniform(0, 1)
conditions = [random_num >= (1 - 0.8062), (random_num < (1 - 0.8062)) & (random_num >= 0.1),
(random_num < 0.1) & (random_num >= 0.05), (random_num < 0.05) &
(random_num >= 0.025), (random_num < 0.025) & (random_num >= 0.0125),
(random_num < 0.0125)]
choices = ['L0', 'L1', 'L2', 'L3', 'L4', 'L5']
L = np.select(conditions, choices)
return L
一个用于df['AL']
。由于您在分配之前使用过df[A]
,因此将其更改为some_number
。
def al_logic(row):
some_number = 1
conditions = [row['L'] == 'L0', row['L'] == 'L1', row['L'] == 'L2', row['L'] == 'L3', row['L'] == 'L4', row['L'] == 'L5']
choices = [some_number * 0.02, some_number * 0.15, some_number * 0.20, some_number * 0.50, some_number * 1, some_number * 1]
AL = np.select(conditions, choices)
return AL
仅在row['Use Type'] =='Commercial Property'
时创建值的逻辑的第三个:
def l_sim(row):
if row['Use Type'] == 'Commercial Property':
if 'L' in row.index:
return al_logic(row)
else:
return l_logic()
else:
return 'NaN'
要启动它:
df['L'] = df.apply(lambda row: l_sim(row), axis=1)
df['AL'] = df.apply(lambda row: l_sim(row), axis=1)
答案 1 :(得分:0)
假设您的数据框至少有两列“ A”和“使用类型”,例如:
df = pd.DataFrame({'Use Type':['Commercial Property']*3+['other']*2, 'A':1})
然后通过以下方式修改您的功能
def l_sim(df,use_type=None):
#check if you want to do it ont he whole datafrmae or a specific Use type
if use_type:
mask = df['Use Type'] == use_type
else:
mask = slice(None)
# generete the random values
df.loc[mask,'RAND'] = np.random.uniform(0, 1, size=df[mask].index.size)
# create conditions (same for both L and AL by the way)
conditions = [ df['RAND'] >= (1 - 0.8062), (df['RAND'] >= 0.1), (df['RAND'] >= 0.05),
(df['RAND'] >= 0.025), (df['RAND'] >= 0.0125), (df['RAND'] < 0.0125)]
#choices for the column L and create the column
choices_L = ['L0', 'L1', 'L2', 'L3', 'L4', 'L5']
df.loc[mask,'L'] = np.select(conditions, choices_L)[mask]
#choices for the column AL and create the column
choices_A = [df['A'] * 0.02, df['A'] * 0.15, df['A'] * 0.20, df['A'] * 0.50,
df['A'] * 1, df['A'] * 1]
df.loc[mask,'AL'] = np.select(conditions, choices_A)[mask]
然后,如果您这样做:
l_sim(df,'Commercial Property')
print (df)
Use Type A RAND L AL
0 Commercial Property 1 0.036593 L3 0.50
1 Commercial Property 1 0.114773 L1 0.15
2 Commercial Property 1 0.651873 L0 0.02
3 other 1 NaN NaN NaN
4 other 1 NaN NaN NaN
和
l_sim(df)
print (df)
Use Type A RAND L AL
0 Commercial Property 1 0.123265 L1 0.15
1 Commercial Property 1 0.906185 L0 0.02
2 Commercial Property 1 0.107588 L1 0.15
3 other 1 0.434560 L0 0.02
4 other 1 0.304901 L0 0.02
我删除了循环for
,因为我看不到要点,并且像上一个问题的answer一样简化了conditions