在迭代过程中为.append（）元素创建新的列表名称

Question

I am trying to create new list_names during an Iteration that I can add Elements to.

# input
df = pd.DataFrame({"R1": [8,2,3], "R2": [-21,-24,4], "R3": [-9,46,6]})

# desired Output
list1 = df.values[0].tolist()
print(list1)
list2 = df.values[1].tolist()
print(list2)
list3 = df.values[2].tolist()
print(list3)

# failed attempt
for i in range(3):
    print(f'{"list"}{i}') = df.values[i].tolist()

Answer 1

You can work on the entire dataframe without iteration. Access the underlying array and convert it into a list of lists in one go.

import pandas as pd
df = pd.DataFrame({"R1": [8,2,3], "R2": [-21,-24,4], "R3": [-9,46,6]})

#out = df.to_numpy().tolist() for pandas >0.24
out = df.values.tolist()
print(out)
print(out[0])
print(out[1])
print(out[2])

Output:

[[8, -21, -9], [2, -24, 46], [3, 4, 6]]
[8, -21, -9]
[2, -24, 46]
[3, 4, 6]

In this manner, you can use the out variable as a collection of all individual lists that you wanted.

---

If you wish to use a dictionary instead, you can also create that as follows:

out_dict = {f"list{i}":lst for i, lst in enumerate(out, start=1)}
#Output:
{'list1': [8, -21, -9], 'list2': [2, -24, 46], 'list3': [3, 4, 6]}

print(out_dict['list1'])
print(out_dict['list2'])
print(out_dict['list3'])

Between the list and dict approaches, you should be able to cover all use-cases you really need. It is generally a bad idea to try making a variable number of variables on the fly. Related read

Answer 2

for i in range(3):
    print('list{} = {}'.format(i, df.loc[i].values.tolist()))

Output

list0 = [8, -21, -9]
list1 = [2, -24, 46]
list2 = [3, 4, 6]

Answer 3

First, to answer your question, to dynamically create variables you can use exec()

for i in range(3):
    exec("List%d=%s" % (i,df.values[i].tolist()))
print(List1)

Secondly, above is a bad practise . This poses a security risk when the values are provided by external data (user input, a file or anything). So it's better avoid using it .

You can always use dictionary in that you can use key value pairs to achieve the same objective (See other answers).