从另一个 Dataframe 的列迭代创建数据帧

Question

Say I have a df:

df = pd.DataFrame({'A.C.1_v': [1, 2, 3], 'B': ['a', 'b', 'c'], 'C.C.1_f': [4, 5, 6], 'D': ['e', 'f', 'g'], 'E': [7, 8, 9]})

Noticed that the col of interest are those nmae includes "C.1_letter"

I have built a list corresponding of selected columns: col_list = [A.C.1_v, C.C.1_f]

Objective: Create several dataframe s as follow (in this illustration only 2 dfs are built, but there could be much more in practice)

The first df

1. Takes the name with the following convention name: "df_AC1_v"
2. Is composed of the values of column A.C.1_v and the values of columns D and E

So, for df_AC1_v we would have the following output: output 1 without iteration

The second df

1. Takes the name with the following convention name: "df_CC1_f"
2. Is composed of the values of column C.C.1_f and the values of columns D and E So, for df_CC1_f, we would have the following output: Output2 without iteration

My point is to do this iteratively, but so far, what I have attempted does not work.

Here are the codes I have done. It bugs in the loop for and I do not understand why. First I extract the col list and create a list as follow:

col_list = list(df)
list_c1 = list(filter(lambda x:'.C.1' in x, col_list))
list_c1 = [str(r) for r in list_c1]

in: list_c1 out:['A.C.1_v', 'C.C.1_f']

Second I isolate the 'C.1'

list_c1_bis = []
for element in list_c1:
    stock = element.split('.C.1')
    list_c1_bis.append(stock)

in: list_c1_bis out:[['A', '_v'], ['C', '_f']]

Until now, I am happy. Where it bugs is the code below:

for line in list_c1_bis:
    name1 ='df'+'_'+line[0]+'C1'+line[1]
    vars()[name1] =  df[[list_c1[0],'D','E']]

My outputs are indeed as follow: in: df_AC1_v ==> OK correct out: output1

in: df_CC1_f ==> Wrong it has taken the inappropriate column A.C.1_v, instead of expected C.C.1_f output2

Your suggestions are welcome !

Thanks a lot for your time and help, that will be truly appreciated

nb: please feel free to modify the first steps that work if you think you have a better solution

Kindest regards

Answer 1

I strongly discouraged you to create variables dynamically with vars , locals or globals . Prefer to use dictionary.

Try

for col in df.columns[df.columns.str.contains(r'[A-Z]\.[0-9]_[a-z]')]:
    name = col.replace('.', '')
    locals()[f"df_{name}"] = df[[col, 'D', 'E']]

Update

If f-strings are not available (Python < 3.6), replace locals()[f"df_{name}"] by locals()["df_{}".format(name)] .

Output:

>>> df_AC1_v
   A.C.1_v  D  E
0        1  e  7
1        2  f  8
2        3  g  9

>>> df_CC1_f
   C.C.1_f  D  E
0        4  e  7
1        5  f  8
2        6  g  9

Alternative with dictionary:

dfs = {}
for col in df.columns[df.columns.str.contains(r'[A-Z]\.[0-9]_[a-z]')]:
    name = col.replace('.', '')
    dfs[name] = df[[col, 'D', 'E']]

Output:

>>> dfs['AC1_v']
   A.C.1_v  D  E
0        1  e  7
1        2  f  8
2        3  g  9

>>> dfs['CC1_f']
   C.C.1_f  D  E
0        4  e  7
1        5  f  8
2        6  g  9

Answer 2

Hi Corralien and first let me thank you for your prompt reply that is truly appreciated.

I have tried the first code

for col in df.columns[df.columns.str.contains(r'[A-Z]\.[0-9]_[a-z]')]:
    name = col.replace('.', '')
    locals()[f"df_{name}"] = df[[col, 'D', 'E']]

But, I have the following error: File "", line 3 locals()[f"df_{name}"] = df[[col, 'D', 'E']] ^ SyntaxError: invalid syntax

I have also tried the second proposed code that gives the solution under dictionary.

dfs = {}
for col in df.columns[df.columns.str.contains(r'[A-Z]\.[0-9]_[a-z]')]:
    name = col.replace('.', '')
    dfs[name] = df[[col, 'D', 'E']]

It runs without error, but when I check the existence of the DFs in: df_AC1_v

I have the following errors: NameError: name 'df_AC1_v' is not defined

I understand that to get the df, it is required to write: dfs['AC1_v']

The second solution is acceptable, but I would prefer the first solution if it worked.

Kindest regards