Python Dictionary Values into Key并追加其他键

Question

我有一个python DataFrame，如下所示:

myDF = pd.DataFrame({"COLUMN_NAME": ["Col1", "Col2", "Col3", "Col4"],
"RULE_1": ["NULL", "DUPLICATE", "TEXT-ONLY", "INTEGER-ONLY"],
"RULE_2": ["DUPLICATE", np.nan, "DUPLICATE", np.nan] })

如何将它转换成这样的字典:

my_dict = {"Col1": ["NULL", "DUPLICATE"], "Col2": ["DUPLICATE"], "Col3": ["TEXT-ONLY", "DUPLICATE"], "Col4": ["INTEGER-ONLY"]}

我被困在做多个循环，但没有真正找到解决方案。

final_rules_dict = defaultdict(list)
for k in rule_dict:
row_dict = rule_dict[k]
for k in row_dict:
col_name = k
if col_name == "COLUMN_NAMES":
final_rules_dict[col_name].append()

Answer 1

这是一个解决方案，适用于您给我们的格式

import json
myDF = pd.DataFrame({"COLUMN_NAME": ["Col1", "Col2", "Col3", "Col4"],
"RULE_1": ["NULL", "DUPLICATE", "TEXT-ONLY", "INTEGER-ONLY"],
"RULE_2": ["DUPLICATE", np.nan, "DUPLICATE", np.nan] })
myDF = myDF.T.reset_index(drop=True)
myDF.columns = myDF.iloc[0]
{
col: [x for x in rows if x is not None] for col, rows in json.loads(pd.io.json.dumps(myDF[1:].to_dict(orient='list'))).items()
}

Answer 2

你可以准备你的空字典:

my_dict = {"Col1": [], "Col2": [], "Col3": [], "Col4": []}

基于"COLUMN_NAME"字段。

然后您可以遍历myDF键(忽略c的"COLUMN_NAME"字段)，并且对于每个键rule_key，您可以编写如下内容:

for i, my_dict_key in enumerate(myDF["COLUMN_NAME"]):
my_dict[my_dict_key].append(myDF[rule_key][i])

Answer 3

d = myDF.set_index('COLUMN_NAME').T.to_dict()

Answer 4

我使用您的方法来获得字典加上一些没有空值的更多转换。

from collections import defaultdict
myDF2 = myDF.T.reset_index(drop=True)
myDF2.columns = myDF2.iloc[0]
myDF3 = myDF2[1:]
dfdict = defaultdict(list)
for (group, col), rule in myDF3.stack().items():
dfdict[col].append(rule)
dfdict