根据字符串中定义的条件填充新列
Fill new column based on conditions defined in a string
我有条件填充在字符串中定义的新列。
condition_string = "colA='yes' & colB='yes' & (colC='yes' | colD='yes'): 'Yes', colA='no' & colB='no' & (colC='no' | colD='no'): 'No', ELSE : 'UNKNOWN'"
字符串可以re-written/structured任何其他格式(字典),然后输入代码得到最终结果。
数据框是
df = pd.DataFrame(
{
'ID': ['AB01', 'AB02', 'AB03', 'AB03', 'AB04','AB05', 'AB06'],
'colA': ["yes","yes",'yes',"no","no",'yes', np.nan],
'colB': [np.nan,'yes','yes',"no",'no', np.nan, "yes"],
'colC': ["yes",'yes', 'yes',"no", "no",np.nan,np.nan],
'colD': ["yes",'no', 'yes',"no",np.nan,"no",np.nan],
}
)
最终结果应该是这样的
如何在不对 condition_string
中的内容进行硬编码的情况下完成此操作。或者你有什么方法可以重组 condition_string
然后应用于数据框?
更新:
如果字典是这样的呢?
condition_string = "colA='yes' & (colB='yes' | colB='no)' &
(colC='yes' | colD='yes'): 'Yes', colA='no' & colB='no' & (colC='no' | colD='no'): 'No', ELSE : 'UNKNOWN'"
数据框就像
df = pd.DataFrame(
{
'ID': ['AB01', 'AB02', 'AB03', 'AB03', 'AB04','AB05', 'AB06'],
'colA': ["yes","yes",'yes',"no","no",'yes', np.nan],
'colB': ["no",'yes','yes',"no",'no', np.nan, "yes"],
'colC': ["yes",'yes', 'yes',"no", "no",np.nan,np.nan],
'colD': ["yes",'no', 'yes',"no",np.nan,"no",np.nan]
}
)
您可以使用 np.where
:
df['results'] = np.where((((df['colA']=='yes') & (df['colB']=='yes')) & ((df['colC']=='yes') | (df['colD']=='yes'))), 'Yes',np.where(((df['colA']=='no') & (df['colB']=='no')) & ((df['colC']=='no' )| (df['colD']=='no')), 'No','UNKNOWN'))
给出:
ID colA colB colC colD decision
0 AB01 yes NaN yes yes UNKNOWN
1 AB02 yes yes yes no Yes
2 AB03 yes yes yes yes Yes
3 AB03 no no no no No
4 AB04 no no no NaN No
5 AB05 yes NaN NaN no UNKNOWN
6 AB06 NaN yes NaN NaN UNKNOWN
IIUC 你想为你的 df
创建任意条件,这可以使用 functools.reduce
和 operator.and_
来完成。然后,您可以使用两个列表(而不是字典)设置条件,第一个是列,第二个是要测试的字符串,最后是 np.select
:
from functools import reduce
from operator import and_
cols = ["colA", "colB", ["colC", "colD"]] # group the cols in a list if they belong to the same group
answer = ["yes", "no"]
conds = [reduce(and_, [df[i].eq(ans) if isinstance(i, str) else df[i].eq(ans).any(1)
for i in cols]) for ans in answer]
df["result"] = np.select(conds, answer, "Unknown")
print (df)
ID colA colB colC colD result
0 AB01 yes NaN yes yes Unknown
1 AB02 yes yes yes no yes
2 AB03 yes yes yes yes yes
3 AB03 no no no no no
4 AB04 no no no NaN no
5 AB05 yes NaN NaN no Unknown
6 AB06 NaN yes NaN NaN Unknown
现在,如果您需要调整条件,只需编辑 cols
和 answer
这两个列表即可。
这里有一个解决方案,可以将您的条件转换为 python 函数,然后将其应用于 DataFrame 的行:
import re
condition_string = "colA='yes' & colB='yes' & (colC='yes' | colD='yes'): 'Yes', colA='no' & colB='no' & (colC='no' | colD='no'): 'No', ELSE : 'UNKNOWN'"
# formatting string as python function apply_cond
for col in df.columns:
condition_string = re.sub(rf"(\W|^){col}(\W|$)", rf"row['{col}']", condition_string)
condition_string = re.sub(rf"row\['{col}'\]\s*=(?!=)", f"row['{col}']==", condition_string)
cond_form = re.sub(r'(:[^[(]+), (?!ELSE)', r'\n\telif ', condition_string) \
.replace(": ", ":\n\t\treturn ") \
.replace("&", "and") \
.replace('|', 'or')
cond_form = re.sub(r", ELSE\s*:", "\n\telse:", cond_form)
function_def = "def apply_cond(row):\n\tif " + cond_form
#print(function_def) # uncomment to see how the function is defined
# executing the function definition of apply_cond
exec(function_def)
# applying the function to each row
df["result"]=df.apply(lambda x: apply_cond(x), axis=1)
print(df)
输出:
ID colA colB colC colD result
0 AB01 yes NaN yes yes UNKNOWN
1 AB02 yes yes yes no Yes
2 AB03 yes yes yes yes Yes
3 AB03 no no no no No
4 AB04 no no no NaN No
5 AB05 yes NaN NaN no UNKNOWN
6 AB06 NaN yes NaN NaN UNKNOWN
您可能希望根据 condition_string
调整字符串格式(我做的很快,可能有一些不受支持的组合)但是如果您自动获取这些字符串,它将避免您重新定义它们。
我有条件填充在字符串中定义的新列。
condition_string = "colA='yes' & colB='yes' & (colC='yes' | colD='yes'): 'Yes', colA='no' & colB='no' & (colC='no' | colD='no'): 'No', ELSE : 'UNKNOWN'"
字符串可以re-written/structured任何其他格式(字典),然后输入代码得到最终结果。
数据框是
df = pd.DataFrame(
{
'ID': ['AB01', 'AB02', 'AB03', 'AB03', 'AB04','AB05', 'AB06'],
'colA': ["yes","yes",'yes',"no","no",'yes', np.nan],
'colB': [np.nan,'yes','yes',"no",'no', np.nan, "yes"],
'colC': ["yes",'yes', 'yes',"no", "no",np.nan,np.nan],
'colD': ["yes",'no', 'yes',"no",np.nan,"no",np.nan],
}
)
最终结果应该是这样的
如何在不对 condition_string
中的内容进行硬编码的情况下完成此操作。或者你有什么方法可以重组 condition_string
然后应用于数据框?
更新: 如果字典是这样的呢?
condition_string = "colA='yes' & (colB='yes' | colB='no)' &
(colC='yes' | colD='yes'): 'Yes', colA='no' & colB='no' & (colC='no' | colD='no'): 'No', ELSE : 'UNKNOWN'"
数据框就像
df = pd.DataFrame(
{
'ID': ['AB01', 'AB02', 'AB03', 'AB03', 'AB04','AB05', 'AB06'],
'colA': ["yes","yes",'yes',"no","no",'yes', np.nan],
'colB': ["no",'yes','yes',"no",'no', np.nan, "yes"],
'colC': ["yes",'yes', 'yes',"no", "no",np.nan,np.nan],
'colD': ["yes",'no', 'yes',"no",np.nan,"no",np.nan]
}
)
您可以使用 np.where
:
df['results'] = np.where((((df['colA']=='yes') & (df['colB']=='yes')) & ((df['colC']=='yes') | (df['colD']=='yes'))), 'Yes',np.where(((df['colA']=='no') & (df['colB']=='no')) & ((df['colC']=='no' )| (df['colD']=='no')), 'No','UNKNOWN'))
给出:
ID colA colB colC colD decision
0 AB01 yes NaN yes yes UNKNOWN
1 AB02 yes yes yes no Yes
2 AB03 yes yes yes yes Yes
3 AB03 no no no no No
4 AB04 no no no NaN No
5 AB05 yes NaN NaN no UNKNOWN
6 AB06 NaN yes NaN NaN UNKNOWN
IIUC 你想为你的 df
创建任意条件,这可以使用 functools.reduce
和 operator.and_
来完成。然后,您可以使用两个列表(而不是字典)设置条件,第一个是列,第二个是要测试的字符串,最后是 np.select
:
from functools import reduce
from operator import and_
cols = ["colA", "colB", ["colC", "colD"]] # group the cols in a list if they belong to the same group
answer = ["yes", "no"]
conds = [reduce(and_, [df[i].eq(ans) if isinstance(i, str) else df[i].eq(ans).any(1)
for i in cols]) for ans in answer]
df["result"] = np.select(conds, answer, "Unknown")
print (df)
ID colA colB colC colD result
0 AB01 yes NaN yes yes Unknown
1 AB02 yes yes yes no yes
2 AB03 yes yes yes yes yes
3 AB03 no no no no no
4 AB04 no no no NaN no
5 AB05 yes NaN NaN no Unknown
6 AB06 NaN yes NaN NaN Unknown
现在,如果您需要调整条件,只需编辑 cols
和 answer
这两个列表即可。
这里有一个解决方案,可以将您的条件转换为 python 函数,然后将其应用于 DataFrame 的行:
import re
condition_string = "colA='yes' & colB='yes' & (colC='yes' | colD='yes'): 'Yes', colA='no' & colB='no' & (colC='no' | colD='no'): 'No', ELSE : 'UNKNOWN'"
# formatting string as python function apply_cond
for col in df.columns:
condition_string = re.sub(rf"(\W|^){col}(\W|$)", rf"row['{col}']", condition_string)
condition_string = re.sub(rf"row\['{col}'\]\s*=(?!=)", f"row['{col}']==", condition_string)
cond_form = re.sub(r'(:[^[(]+), (?!ELSE)', r'\n\telif ', condition_string) \
.replace(": ", ":\n\t\treturn ") \
.replace("&", "and") \
.replace('|', 'or')
cond_form = re.sub(r", ELSE\s*:", "\n\telse:", cond_form)
function_def = "def apply_cond(row):\n\tif " + cond_form
#print(function_def) # uncomment to see how the function is defined
# executing the function definition of apply_cond
exec(function_def)
# applying the function to each row
df["result"]=df.apply(lambda x: apply_cond(x), axis=1)
print(df)
输出:
ID colA colB colC colD result
0 AB01 yes NaN yes yes UNKNOWN
1 AB02 yes yes yes no Yes
2 AB03 yes yes yes yes Yes
3 AB03 no no no no No
4 AB04 no no no NaN No
5 AB05 yes NaN NaN no UNKNOWN
6 AB06 NaN yes NaN NaN UNKNOWN
您可能希望根据 condition_string
调整字符串格式(我做的很快,可能有一些不受支持的组合)但是如果您自动获取这些字符串,它将避免您重新定义它们。