Creating a Dataframe from a Dictionary-CodePudding

I thought that I had a very good grasp of pandas data frames, but for some reason, I can convert this dictionary to a dataframe - I'm going to post the full code with constant variables in case someone wants to tag alone:

import pandas as pd
import numpy as np

patches = 26
min_sl = 2.1
max_sl = 104.1

array = np.arange(min_sl, max_sl   1, step=1)

##Creating an array List
array_list = [i for i in array]

## Creating a dictionary
dict_l={i:"" for i in range(1,patches 1)}

## Populating the dictionary
start = 0
end = 4
for i in dict_l:
    dict_l[i] = array_list[start:end]
    start = start   5
    end = end   5

This creates the following dictionary

{1: [2.1, 3.1, 4.1, 5.1],
 2: [7.1, 8.1, 9.1, 10.1],
 3: [12.1, 13.1, 14.1, 15.1],
 4: [17.1, 18.1, 19.1, 20.1],
 5: [22.1, 23.1, 24.1, 25.1],
 6: [27.1, 28.1, 29.1, 30.1],
 7: [32.1, 33.1, 34.1, 35.1],
 8: [37.1, 38.1, 39.1, 40.1],
 9: [42.1, 43.1, 44.1, 45.1],
 10: [47.1, 48.1, 49.1, 50.1],
 11: [52.1, 53.1, 54.1, 55.1],
 12: [57.1, 58.1, 59.1, 60.1],
 13: [62.1, 63.1, 64.1, 65.1],
 14: [67.1, 68.1, 69.1, 70.1],
 15: [72.1, 73.1, 74.1, 75.1],
 16: [77.1, 78.1, 79.1, 80.1],
 17: [82.1, 83.1, 84.1, 85.1],
 18: [87.1, 88.1, 89.1, 90.1],
 19: [92.1, 93.1, 94.1, 95.1],
 20: [97.1, 98.1, 99.1, 100.1],
 21: [102.1, 103.1, 104.1],
 22: [],
 23: [],
 24: [],
 25: [],
 26: []}

What I want is a simple dataframe that puts each list in to a row, so for example index 1 will have a list with this values : [2.1, 3.1, 4.1, 5.1],

result = pd.DataFrame.from_dict(dict_l, orient='index')
result

As you can see my result is not what I want, because I want just one column

If I don't specify the function from_dict, I get an error, so I'm wondering if anyone has any idea on how to do this..

I tried to concatenate all columns with a comma, which is in theory what I would like to do:

result['Line Name'] = result[[i for i in range(0,4,1)]].apply(lambda x: ', '.join(x[x.notnull()]), axis=1)

However, I get a type error TypeError: sequence item 0: expected str instance, float found

Does anyone know how to accomplish this?

CodePudding user response：

df = pd.DataFrame({k:[v] for k,v in dict_l.items()}).astype(str)
df = df.stack().reset_index(drop=True)
print(df)

output:

0          [2.1, 3.1, 4.1, 5.1]
1         [7.1, 8.1, 9.1, 10.1]
2      [12.1, 13.1, 14.1, 15.1]
3      [17.1, 18.1, 19.1, 20.1]
4      [22.1, 23.1, 24.1, 25.1]
5      [27.1, 28.1, 29.1, 30.1]
6      [32.1, 33.1, 34.1, 35.1]
7      [37.1, 38.1, 39.1, 40.1]
8      [42.1, 43.1, 44.1, 45.1]
9      [47.1, 48.1, 49.1, 50.1]
10     [52.1, 53.1, 54.1, 55.1]
11     [57.1, 58.1, 59.1, 60.1]
12     [62.1, 63.1, 64.1, 65.1]
13     [67.1, 68.1, 69.1, 70.1]
14     [72.1, 73.1, 74.1, 75.1]
15     [77.1, 78.1, 79.1, 80.1]
16     [82.1, 83.1, 84.1, 85.1]
17     [87.1, 88.1, 89.1, 90.1]
18     [92.1, 93.1, 94.1, 95.1]
19    [97.1, 98.1, 99.1, 100.1]
20        [102.1, 103.1, 104.1]
21                           []
22                           []
23                           []
24                           []
25                           []

CodePudding user response：

Just put the list in the list:

...
    dict_l[i] = [array_list[start:end]]
...

Prints:

                            0
1        [2.1, 3.1, 4.1, 5.1]
2       [7.1, 8.1, 9.1, 10.1]
3    [12.1, 13.1, 14.1, 15.1]
4    [17.1, 18.1, 19.1, 20.1]
5    [22.1, 23.1, 24.1, 25.1]
6    [27.1, 28.1, 29.1, 30.1]
7    [32.1, 33.1, 34.1, 35.1]
8    [37.1, 38.1, 39.1, 40.1]
9    [42.1, 43.1, 44.1, 45.1]
10   [47.1, 48.1, 49.1, 50.1]
11   [52.1, 53.1, 54.1, 55.1]
12   [57.1, 58.1, 59.1, 60.1]
13   [62.1, 63.1, 64.1, 65.1]
14   [67.1, 68.1, 69.1, 70.1]
15   [72.1, 73.1, 74.1, 75.1]
16   [77.1, 78.1, 79.1, 80.1]
17   [82.1, 83.1, 84.1, 85.1]
18   [87.1, 88.1, 89.1, 90.1]
19   [92.1, 93.1, 94.1, 95.1]
20  [97.1, 98.1, 99.1, 100.1]
21      [102.1, 103.1, 104.1]
22                         []
23                         []
24                         []
25                         []
26                         []

CodePudding user response：

DataFrames are two-dimensional data structures, so the lists of the dictionary are interpreted as rows. Either construct a Series (1D) and convert it to a DataFrame

result = pd.Series(dict_l).to_frame()

or wrap the dictionary values with extra brackets to create a "3rd dimension".

result = pd.DataFrame([[lst] for lst in dict_l.values()], 
                      index=dict_l.keys())

Output:

>>> result

                            0
1        [2.1, 3.1, 4.1, 5.1]
2       [7.1, 8.1, 9.1, 10.1]
3    [12.1, 13.1, 14.1, 15.1]
4    [17.1, 18.1, 19.1, 20.1]
5    [22.1, 23.1, 24.1, 25.1]
6    [27.1, 28.1, 29.1, 30.1]
7    [32.1, 33.1, 34.1, 35.1]
8    [37.1, 38.1, 39.1, 40.1]
9    [42.1, 43.1, 44.1, 45.1]
10   [47.1, 48.1, 49.1, 50.1]
11   [52.1, 53.1, 54.1, 55.1]
12   [57.1, 58.1, 59.1, 60.1]
13   [62.1, 63.1, 64.1, 65.1]
14   [67.1, 68.1, 69.1, 70.1]
15   [72.1, 73.1, 74.1, 75.1]
16   [77.1, 78.1, 79.1, 80.1]
17   [82.1, 83.1, 84.1, 85.1]
18   [87.1, 88.1, 89.1, 90.1]
19   [92.1, 93.1, 94.1, 95.1]
20  [97.1, 98.1, 99.1, 100.1]
21      [102.1, 103.1, 104.1]
22                         []
23                         []
24                         []
25                         []
26                         []