Pandas plotting two columns with series defined by value in third column - python

Hi I have a pandas dataframe that looks like
deflector wFlow aContent DO Difference
64 3 127.5 10 2.007395
65 3 127.5 3 1.163951
66 3 127.5 5 1.451337
67 3 127.5 7 1.535639
68 3 24.0 10 1.046328
69 3 24.0 3 0.854763
70 3 24.0 5 0.766780
71 3 24.0 7 0.905270
72 3 56.0 10 1.274954
73 3 56.0 3 1.298657
74 3 56.0 5 1.049621
75 3 56.0 7 1.004255
76 3 88.0 10 1.194174
77 3 88.0 3 1.056968
78 3 88.0 5 1.066173
79 3 88.0 7 1.097231
I would like to plot the aContent column vs the DO Difference column with each line defined by the wFlow column (x = aContent, y = DO Difference, 4 different lines, one for each wFlow.
Thanks!

You can pivot the data and use pandas.dataframe.plot:
df.pivot(index='aContent',columns='wFlow',values='DO Difference').plot()

Related

Swipe or turn data for stacked bar chart in Matplotlib

I'm trying to create or generate some graphs in stacked bar I'm using this data:
index 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
0 No 94 123 96 108 122 106.0 95.0 124 104 118 73 82 106 124 109 70 59
1 Yes 34 4 33 21 5 25.0 34.0 5 21 9 55 46 21 3 19 59 41
2 Dont know 1 2 1 1 2 NaN NaN 1 4 2 2 2 2 2 2 1 7
Basically I want to use the columns names as x and the Yes, No, Don't know as the Y values, here is my code and the result that I have at the moment.
ax = dfu.plot.bar(x='index', stacked=True)
UPDATE:
Here is an example:
data = [{0:1,1:2,2:3},{0:3,1:2,2:1},{0:1,1:1,2:1}]
index = ["yes","no","dont know"]
df = pd.DataFrame(data,index=index)
df.T.plot.bar(stacked=True) # Note .T is used to transpose the DataFrame

How to perform operations with columns from different datasets with different indexation?

The goal
A bit of background, to get familiar with variables and understand what the problem is:
floor, square, matc and volume are tables or dataframes, all share same column "id" (which simply goes from 1 to 100), so every row is unique;
floor and square also share column "room_name";
volume is generally equivalent to floor, except all rows with rooms ("room_name") that have no values in "square" column of square dataframe were dropped; This implies that some values of "id" are missing
That done, I needed to create a new column in volume dataframe, which would consist of multiplication of one of its own columns with two other columns from matc and square dataframes.
The problem
This seemingly simple interaction turned out to be quite difficult, because, well, the columns I am working with are of different length (except for square and matc, they are the same) and I need to align them by "id". To make matters worse, when called directly as volume['coefLoosening'] (please note that coefLoosening does not originate from floor and is added after the table is created), it returns a series with its own index and no way to relate it to "id".
What I tried
Whilst trying to solve the issue, I came up with this abomination:
volume = volume.merge(pd.DataFrame({"id": matc.loc[matc["id"].isin(volume["id"])]["id"], "tempCoef": volume['coefLoosening'] * matc.loc[matc["id"].isin(volume["id"])]['width'] * square.loc[square["id"].isin(volume["id"])]['square']}), how = "left", on = ["id"])
This, however, misaligns "id" column completely, somehow creating more rows. For instance, this what `` returns:
index id tempCoef
0 1.0 960.430612244898
1 2.0 4665.499999999999
2 NaN NaN
3 4.0 2425.44652173913
4 5.0 5764.964210526316
5 6.0 55201.68727272727
6 NaN NaN
7 NaN NaN
8 NaN NaN
9 10.0 1780.7208791208789
10 11.0 6075.385074626865
11 12.0 10400.94
12 13.0 31.378285714285713
13 NaN NaN
14 NaN NaN
15 NaN NaN
16 17.0 10505.431451612903
17 18.0 1208.994845360825
18 NaN NaN
19 NaN NaN
20 21.0 568.8900000000001
21 22.0 4275.416470588235
22 NaN NaN
23 NaN NaN
24 25.0 547.04
25 26.0 2090.666111111111
26 27.0 2096.88406779661
27 NaN NaN
28 29.0 8324.566547619048
29 NaN NaN
30 NaN NaN
31 NaN NaN
32 33.0 2459.8314736842103
33 34.0 2177.778461538461
34 35.0 166.1257142857143
35 36.0 1866.8492307692304
36 37.0 3598.1470588235293
37 38.0 21821.709411764703
38 NaN NaN
39 40.0 2999.248
40 41.0 980.3136
41 42.0 2641.3503947368426
42 NaN NaN
43 44.0 25829.878148148146
44 45.0 649.3632
45 46.0 10895.386666666667
46 NaN NaN
47 NaN NaN
48 49.0 825.9879310344828
49 50.0 15951.941666666671
50 51.0 2614.9343434343436
51 52.0 2462.30625
52 NaN NaN
53 NaN NaN
54 55.0 1366.8287671232877
55 56.0 307.38
56 57.0 11601.975
57 58.0 1002.5415730337081
58 59.0 2493.4532432432434
59 60.0 981.7482608695652
61 62.0 NaN
63 64.0 NaN
65 66.0 NaN
67 68.0 NaN
73 74.0 NaN
75 76.0 NaN
76 77.0 NaN
77 78.0 NaN
78 79.0 NaN
80 81.0 NaN
82 83.0 NaN
84 85.0 NaN
88 89.0 NaN
89 90.0 NaN
90 91.0 NaN
92 93.0 NaN
94 95.0 NaN
95 96.0 NaN
97 98.0 NaN
98 99.0 NaN
99 100.0 NaN
For clarity, no values in any of columns in the operation have NaNs in them.
This is what 'volume["coefLoosening"]` returns:
0 1.020408
1 1.515152
2 2.000000
3 4.347826
4 5.263158
5 9.090909
6 1.162791
7 1.149425
8 1.851852
9 1.098901
10 1.492537
11 2.083333
12 1.428571
13 1.010101
14 1.562500
15 3.448276
16 1.612903
17 1.030928
18 33.333333
19 1.000000
20 1.123596
21 1.960784
22 2.127660
23 2.857143
24 1.369863
25 1.111111
26 1.694915
27 1.492537
28 1.190476
29 1.818182
30 1.612903
31 12.500000
32 1.052632
33 3.846154
34 2.040816
35 1.098901
36 2.941176
37 2.941176
38 2.857143
39 1.111111
40 1.333333
41 1.315789
42 3.703704
43 3.703704
44 2.000000
45 33.333333
46 12.500000
47 1.149425
48 1.724138
49 4.166667
50 1.010101
51 1.041667
52 1.162791
53 3.225806
54 1.369863
55 1.666667
56 4.545455
57 1.123596
58 1.351351
59 2.173913
and finally, this is what volume["id"] returns (to compare to the result of «abomination»):
0 1
1 2
2 4
3 5
4 6
5 10
6 11
7 12
8 13
9 17
10 18
11 21
12 22
13 25
14 26
15 27
16 29
17 33
18 34
19 35
20 36
21 37
22 38
23 40
24 41
25 42
26 44
27 45
28 46
29 49
30 50
31 51
32 52
33 55
34 56
35 57
36 58
37 59
38 60
39 62
40 64
41 66
42 68
43 74
44 76
45 77
46 78
47 79
48 81
49 83
50 85
51 89
52 90
53 91
54 93
55 95
56 96
57 98
58 99
59 100
Some thoughts
I believe, part of the problem is how pandas returns columns (as series with default indexation) and I don't know how to work around that.
Another source of the problem might be the way how .loc() method returns its result. In the case of matc.loc[matc["id"].isin(volume["id"])]['width'] it is:
0 15.98
1 36.12
3 32.19
4 18.54
5 98.96
9 64.56
10 58.20
11 55.08
12 3.84
16 77.31
17 15.25
20 63.21
21 76.32
24 10.52
25 54.65
26 95.46
28 79.67
32 57.01
33 27.54
34 7.36
35 36.44
36 23.64
37 78.98
39 92.19
40 31.26
41 61.71
43 70.07
44 10.91
45 4.24
48 7.35
49 46.70
50 97.69
51 32.03
54 13.50
55 42.30
56 94.71
57 37.49
58 57.86
59 50.29
61 18.18
63 88.26
65 4.28
67 28.89
73 4.05
75 22.37
76 52.20
77 98.29
78 72.98
80 6.07
82 35.80
84 64.16
88 23.60
89 45.05
90 21.14
92 31.21
94 46.04
95 7.15
97 27.70
98 31.93
99 79.62
which is shifted by -1 and I don't see a way to change this manually.
So, any ideas? Maybe there is answered analogue of this question (because I tried to search it before asking, but found nothing)?
Data
Minimal columns of tables required to replicate this (because stack overflow does not allow files to be uploaded)
volume:
index,id,room_name,coefLoosening
0,1,6,1.0204081632653061
1,2,7,1.5151515151515151
2,4,3,2.0
3,5,7,4.3478260869565215
4,6,4,5.2631578947368425
5,10,7,9.090909090909092
6,11,5,1.1627906976744187
7,12,4,1.1494252873563218
8,13,1,1.8518518518518516
9,17,3,1.0989010989010988
10,18,3,1.4925373134328357
11,21,3,2.0833333333333335
12,22,7,1.4285714285714286
13,25,3,1.0101010101010102
14,26,6,1.5625
15,27,6,3.4482758620689657
16,29,4,1.6129032258064517
17,33,2,1.0309278350515465
18,34,2,33.333333333333336
19,35,5,1.0
20,36,4,1.1235955056179776
21,37,2,1.9607843137254901
22,38,6,2.127659574468085
23,40,5,2.857142857142857
24,41,6,1.36986301369863
25,42,3,1.1111111111111112
26,44,2,1.6949152542372883
27,45,4,1.4925373134328357
28,46,2,1.1904761904761905
29,49,5,1.8181818181818181
30,50,4,1.6129032258064517
31,51,2,12.5
32,52,3,1.0526315789473684
33,55,6,3.846153846153846
34,56,5,2.0408163265306123
35,57,5,1.0989010989010988
36,58,4,2.941176470588235
37,59,5,2.941176470588235
38,60,5,2.857142857142857
39,62,7,1.1111111111111112
40,64,7,1.3333333333333333
41,66,7,1.3157894736842106
42,68,3,3.7037037037037033
43,74,5,3.7037037037037033
44,76,4,2.0
45,77,3,33.333333333333336
46,78,4,12.5
47,79,5,1.1494252873563218
48,81,5,1.7241379310344829
49,83,4,4.166666666666667
50,85,2,1.0101010101010102
51,89,4,1.0416666666666667
52,90,1,1.1627906976744187
53,91,2,3.2258064516129035
54,93,2,1.36986301369863
55,95,1,1.6666666666666667
56,96,4,4.545454545454546
57,98,7,1.1235955056179776
58,99,7,1.3513513513513513
59,100,5,2.1739130434782608
matc:
index,id,width
0,1,15.98
1,2,36.12
2,3,63.41
3,4,32.19
4,5,18.54
5,6,98.96
6,7,5.65
7,8,97.42
8,9,50.88
9,10,64.56
10,11,58.2
11,12,55.08
12,13,3.84
13,14,75.87
14,15,96.51
15,16,42.08
16,17,77.31
17,18,15.25
18,19,81.43
19,20,98.71
20,21,63.21
21,22,76.32
22,23,22.59
23,24,30.79
24,25,10.52
25,26,54.65
26,27,95.46
27,28,49.93
28,29,79.67
29,30,45.0
30,31,59.14
31,32,62.25
32,33,57.01
33,34,27.54
34,35,7.36
35,36,36.44
36,37,23.64
37,38,78.98
38,39,47.8
39,40,92.19
40,41,31.26
41,42,61.71
42,43,93.11
43,44,70.07
44,45,10.91
45,46,4.24
46,47,35.39
47,48,99.1
48,49,7.35
49,50,46.7
50,51,97.69
51,52,32.03
52,53,48.61
53,54,33.44
54,55,13.5
55,56,42.3
56,57,94.71
57,58,37.49
58,59,57.86
59,60,50.29
60,61,77.98
61,62,18.18
62,63,3.42
63,64,88.26
64,65,48.66
65,66,4.28
66,67,20.78
67,68,28.89
68,69,27.17
69,70,57.48
70,71,59.07
71,72,12.63
72,73,22.06
73,74,4.05
74,75,22.3
75,76,22.37
76,77,52.2
77,78,98.29
78,79,72.98
79,80,49.37
80,81,6.07
81,82,28.85
82,83,35.8
83,84,66.74
84,85,64.16
85,86,33.64
86,87,66.36
87,88,34.51
88,89,23.6
89,90,45.05
90,91,21.14
91,92,97.27
92,93,31.21
93,94,13.04
94,95,46.04
95,96,7.15
96,97,47.87
97,98,27.7
98,99,31.93
99,100,79.62
square:
index,id,room_name,square
0,1,5,58.9
1,2,3,85.25
2,3,5,90.39
3,4,3,17.33
4,5,2,59.08
5,6,4,61.36
6,7,2,29.02
7,8,2,59.63
8,9,6,98.31
9,10,4,25.1
10,11,3,69.94
11,12,7,90.64
12,13,4,5.72
13,14,6,29.96
14,15,4,59.06
15,16,1,41.85
16,17,7,84.25
17,18,4,76.9
18,19,1,17.2
19,20,4,60.9
20,21,1,8.01
21,22,2,28.57
22,23,1,65.07
23,24,1,20.24
24,25,7,37.96
25,26,7,34.43
26,27,3,12.96
27,28,6,80.96
28,29,5,87.77
29,30,2,95.67
30,31,1,10.4
31,32,1,30.96
32,33,6,40.99
33,34,7,20.56
34,35,5,11.06
35,36,4,46.62
36,37,3,51.75
37,38,4,93.94
38,39,5,62.64
39,40,6,29.28
40,41,3,23.52
41,42,6,32.53
42,43,1,33.3
43,44,3,99.53
44,45,5,29.76
45,46,7,77.09
46,47,1,71.31
47,48,2,59.22
48,49,1,65.18
49,50,7,81.98
50,51,7,26.5
51,52,3,73.8
52,53,6,78.52
53,54,6,69.67
54,55,6,73.91
55,56,6,4.36
56,57,5,26.95
57,58,2,23.8
58,59,2,31.89
59,60,1,8.98
60,61,1,88.76
61,62,5,88.75
62,63,4,44.94
63,64,4,81.13
64,65,5,48.39
65,66,3,55.63
66,67,7,46.28
67,68,3,40.85
68,69,7,54.37
69,70,3,14.01
70,71,6,20.13
71,72,2,90.67
72,73,3,4.28
73,74,4,56.18
74,75,3,74.8
75,76,5,10.34
76,77,6,15.94
77,78,2,29.4
78,79,6,60.8
79,80,3,13.05
80,81,3,49.46
81,82,1,75.76
82,83,1,84.27
83,84,5,76.36
84,85,3,75.98
85,86,7,77.81
86,87,2,56.34
87,88,1,43.93
88,89,5,30.64
89,90,5,55.78
90,91,5,88.26
91,92,6,15.11
92,93,1,20.64
93,94,2,5.08
94,95,1,82.31
95,96,4,76.92
96,97,1,53.47
97,98,2,2.7
98,99,7,77.12
99,100,4,29.43
floor:
index,id,room_name
0,1,6
1,2,7
2,3,12
3,4,3
4,5,7
5,6,4
6,7,8
7,8,11
8,9,10
9,10,7
10,11,5
11,12,4
12,13,1
13,14,11
14,15,12
15,16,9
16,17,3
17,18,3
18,19,9
19,20,12
20,21,3
21,22,7
22,23,8
23,24,12
24,25,3
25,26,6
26,27,6
27,28,10
28,29,4
29,30,10
30,31,9
31,32,11
32,33,2
33,34,2
34,35,5
35,36,4
36,37,2
37,38,6
38,39,11
39,40,5
40,41,6
41,42,3
42,43,11
43,44,2
44,45,4
45,46,2
46,47,9
47,48,12
48,49,5
49,50,4
50,51,2
51,52,3
52,53,9
53,54,10
54,55,6
55,56,5
56,57,5
57,58,4
58,59,5
59,60,5
60,61,12
61,62,7
62,63,12
63,64,7
64,65,11
65,66,7
66,67,12
67,68,3
68,69,8
69,70,11
70,71,12
71,72,8
72,73,12
73,74,5
74,75,11
75,76,4
76,77,3
77,78,4
78,79,5
79,80,12
80,81,5
81,82,12
82,83,4
83,84,8
84,85,2
85,86,8
86,87,8
87,88,9
88,89,4
89,90,1
90,91,2
91,92,9
92,93,2
93,94,12
94,95,1
95,96,4
96,97,8
97,98,7
98,99,7
99,100,5
IIUC you overcomplicated things. The whole thing about merging on id is that you don't need to filter the other df's beforehand on id with loc and isin like you tried to do, merge will do that for you.
You could multiply square and width at the square_df (matc_df would also work since they have same length and id).
Then merge this new column to the volume_df (which filters the multiplied result only to the id's which are found in the volume_df) and multiply it again.
square_df['square*width'] = square_df['square'] * matc_df['width']
df = volume_df.merge(square_df[['id', 'square*width']], on='id', how='left')
df['result'] = df['coefLoosening'] * df['square*width']
Output df:
id room_name coefLoosening square*width result
0 1 6 1.020408 941.2220 960.430612
1 2 7 1.515152 3079.2300 4665.500000
2 4 3 2.000000 557.8527 1115.705400
3 5 7 4.347826 1095.3432 4762.361739
4 6 4 5.263158 6072.1856 31958.871579
5 10 7 9.090909 1620.4560 14731.418182
6 11 5 1.162791 4070.5080 4733.148837
7 12 4 1.149425 4992.4512 5738.449655
8 13 1 1.851852 21.9648 40.675556
9 17 3 1.098901 6513.3675 7157.546703
10 18 3 1.492537 1172.7250 1750.335821
11 21 3 2.083333 506.3121 1054.816875
12 22 7 1.428571 2180.4624 3114.946286
13 25 3 1.010101 399.3392 403.372929
14 26 6 1.562500 1881.5995 2939.999219
15 27 6 3.448276 1237.1616 4266.074483
16 29 4 1.612903 6992.6359 11278.445000
17 33 2 1.030928 2336.8399 2409.113299
18 34 2 33.333333 566.2224 18874.080000
19 35 5 1.000000 81.4016 81.401600
20 36 4 1.123596 1698.8328 1908.800899
21 37 2 1.960784 1223.3700 2398.764706
22 38 6 2.127660 7419.3812 15785.917447
23 40 5 2.857143 2699.3232 7712.352000
24 41 6 1.369863 735.2352 1007.171507
25 42 3 1.111111 2007.4263 2230.473667
26 44 2 1.694915 6974.0671 11820.452712
27 45 4 1.492537 324.6816 484.599403
28 46 2 1.190476 326.8616 389.120952
29 49 5 1.818182 479.0730 871.041818
30 50 4 1.612903 3828.4660 6174.945161
31 51 2 12.500000 2588.7850 32359.812500
32 52 3 1.052632 2363.8140 2488.225263
33 55 6 3.846154 997.7850 3837.634615
34 56 5 2.040816 184.4280 376.383673
35 57 5 1.098901 2552.4345 2804.873077
36 58 4 2.941176 892.2620 2624.300000
37 59 5 2.941176 1845.1554 5426.927647
38 60 5 2.857143 451.6042 1290.297714
39 62 7 1.111111 1613.4750 1792.750000
40 64 7 1.333333 7160.5338 9547.378400
41 66 7 1.315789 238.0964 313.284737
42 68 3 3.703704 1180.1565 4370.950000
43 74 5 3.703704 227.5290 842.700000
44 76 4 2.000000 231.3058 462.611600
45 77 3 33.333333 832.0680 27735.600000
46 78 4 12.500000 2889.7260 36121.575000
47 79 5 1.149425 4437.1840 5100.211494
48 81 5 1.724138 300.2222 517.624483
49 83 4 4.166667 3016.8660 12570.275000
50 85 2 1.010101 4874.8768 4924.117980
51 89 4 1.041667 723.1040 753.233333
52 90 1 1.162791 2512.8890 2921.963953
53 91 2 3.225806 1865.8164 6018.762581
54 93 2 1.369863 644.1744 882.430685
55 95 1 1.666667 3789.5524 6315.920667
56 96 4 4.545455 549.9780 2499.900000
57 98 7 1.123596 74.7900 84.033708
58 99 7 1.351351 2462.4416 3327.623784
59 100 5 2.173913 2343.2166 5093.949130

How to create a rolling window in pandas with another condition

I have a data frame with 2 columns
df = pd.DataFrame(np.random.randint(0,100,size=(100, 2)), columns=list('AB'))
A B
0 11 10
1 61 30
2 24 54
3 47 52
4 72 42
... ... ...
95 61 2
96 67 41
97 95 30
98 29 66
99 49 22
100 rows × 2 columns
Now I want to create a third column, which is a rolling window max of col 'A' BUT
the max has to be lower than the corresponding value in col 'B'. In other words I want the value of the 4 (using a window size of 4) in column 'A' closest to the value in col 'B', yet smaller than B
So for example in row
3 47 52
the new value I am looking for, is not 61 but 47, because it is the highest value of the 4 that is not higher than 52
pseudo code
df['C'] = df['A'].rolling(window=4).max() where < df['B']
You can use concat + shift to create a wide DataFrame with the previous values, which makes complicated rolling calculations a bit easier.
Sample Data
np.random.seed(42)
df = pd.DataFrame(np.random.randint(0, 100, size=(100, 2)), columns=list('AB'))
Code
N = 4
# End slice ensures same default min_periods behavior to `.rolling`
df1 = pd.concat([df['A'].shift(i).rename(i) for i in range(N)], axis=1).iloc[N-1:]
# Remove values larger than B, then find the max of remaining.
df['C'] = df1.where(df1.lt(df.B, axis=0)).max(1)
print(df.head(15))
A B C
0 51 92 NaN # Missing b/c min_periods
1 14 71 NaN # Missing b/c min_periods
2 60 20 NaN # Missing b/c min_periods
3 82 86 82.0
4 74 74 60.0
5 87 99 87.0
6 23 2 NaN # Missing b/c 82, 74, 87, 23 all > 2
7 21 52 23.0 # Max of 21, 23, 87, 74 which is < 52
8 1 87 23.0
9 29 37 29.0
10 1 63 29.0
11 59 20 1.0
12 32 75 59.0
13 57 21 1.0
14 88 48 32.0
You can use a custom function to .apply to the rolling window. In this case, you can use a default argument to pass in the B column.
df = pd.DataFrame(np.random.randint(0,100,size=(100, 2)), columns=('AB'))
def rollup(a, B=df.B):
ix = a.index.max()
b = B[ix]
return a[a<b].max()
df['C'] = df.A.rolling(4).apply(rollup)
df
# returns:
A B C
0 8 17 NaN
1 23 84 NaN
2 75 84 NaN
3 86 24 23.0
4 52 83 75.0
.. .. .. ...
95 38 22 NaN
96 53 48 38.0
97 45 4 NaN
98 3 92 53.0
99 91 86 53.0
The NaN values occur when no number in the window of A is less than B or at the start of the series when the window is too big for the first few rows.
You can use where to replace values that don't fulfill the condition with np.nan and then use rolling(window=4, min_periods=1):
In [37]: df['C'] = df['A'].where(df['A'] < df['B'], np.nan).rolling(window=4, min_periods=1).max()
In [38]: df
Out[38]:
A B C
0 0 1 0.0
1 1 2 1.0
2 2 3 2.0
3 10 4 2.0
4 4 5 4.0
5 5 6 5.0
6 10 7 5.0
7 10 8 5.0
8 10 9 5.0
9 10 10 NaN

dataframe concatenating with indexing

I have a Python dataframe that reads from a file
the next step I do is to break dataset into 2 datasets df_LastYear & df_ThisYear
Note : that Index is not continuous missing 2 & 6
ID AdmissionAge
0 14 68
1 22 86
3 78 40
4 124 45
5 128 35
7 148 92
8 183 71
9 185 98
10 219 79
after applying some predictive models I get results of predictive values y_ThisYear
Prediction
0 2.400000e+01
1 1.400000e+01
2 1.000000e+00
3 2.096032e+09
4 2.000000e+00
5 -7.395179e+11
6 6.159412e+06
7 5.592327e+07
8 5.303477e+08
9 5.500000e+00
10 6.500000e+00
I am trying to concat both datasets df_ThisYear and y_ThisYear into one dataset
but I always get these results
ID AdmissionAge Prediction
0 14.0 68.0 2.400000e+01
1 22.0 86.0 1.400000e+01
2 NaN NaN 1.000000e+00
3 78.0 40.0 2.096032e+09
4 124.0 45.0 2.000000e+00
5 128.0 35.0 -7.395179e+11
6 NaN NaN 6.159412e+06
7 148.0 92.0 5.592327e+07
8 183.0 71.0 5.303477e+08
9 185.0 98.0 5.500000e+00
10 219.0 79.0 6.500000e+00
There are NaNs which did not exist before
I found that these NaNs are belonging to the index which was not included in df_ThisYear
Therefore I try reset index so I get continuous Indices
I used
df_ThisYear.reset_index(drop=True)
but still getting same indices
How to fix this problem so I can concatenate df_ThisYear with y_ThisYear correctly?
Then you just need join
df.join(Y)
ID AdmissionAge Prediction
0 14 68 2.400000e+01
1 22 86 1.400000e+01
3 78 40 2.096032e+09
4 124 45 2.000000e+00
5 128 35 -7.395179e+11
7 148 92 5.592327e+07
8 183 71 5.303477e+08
9 185 98 5.500000e+00
10 219 79 6.500000e+00
If you are really excited about using concat, you can provide 'inner' to the how argument:
pd.concat([df_ThisYear, y_ThisYear], axis=1, join='inner')
This returns
Out[6]:
ID AdmissionAge Prediction
0 14 68 2.400000e+01
1 22 86 1.400000e+01
3 78 40 2.096032e+09
4 124 45 2.000000e+00
5 128 35 -7.395179e+11
7 148 92 5.592327e+07
8 183 71 5.303477e+08
9 185 98 5.500000e+00
10 219 79 6.500000e+00
Because y_ThisYear has different index than df_ThisYear
When I joined both using
df_ThisYear.join(y_ThisYear )
it started to match each number it its matching index
I know this is right if indices are actually represent the same record i.e. index 7 in df_ThisYear value is matching y_ThisYear index 7 too
In my case I just want to match first record in y_ThisYear to the first in df_ThisYear regardless of their index number
I found this code that does that.
df_ThisYear = pd.concat([df_ThisYear.reset_index(drop=True), pd.DataFrame(y_ThisYear)], axis=1)
Thanks for everyone helped with the answer

Python Pandas self merge on previous data

I have a DataFrame that contains many years worth of data. I want to make a couple columns containing the previous years' data from the same DataFrame. Here's an example:
df = pd.DataFrame({'id': [1,1,1,2,2,2,3,4,5,3,3,3,4],
'yr': [87,88,89,54,55,53,87,87,89,90,91,92,86],
'data': '1-87 1-88 1-89 2-54 2-55 2-53 3-87 4-87 5-89 3-90 3-91 3-92 4-86'.split()})
data id yr
0 1-87 1 87
1 1-88 1 88
2 1-89 1 89
3 2-54 2 54
4 2-55 2 55
5 2-53 2 53
6 3-87 3 87
7 4-87 4 87
8 5-89 5 89
9 3-90 3 90
10 3-91 3 91
11 3-92 3 92
12 4-86 4 86
I'd like to add on another column that shows the previous years' data for that id number. like this:
data id yr last_year_data
0 1-87 1 87 NaN
1 1-88 1 88 1-87
2 1-89 1 89 1-88
3 2-54 2 54 2-53
4 2-55 2 55 2-54
5 2-53 2 53 NaN
6 3-87 3 87 NaN
7 4-87 4 87 4-86
8 5-89 5 89 NaN
9 3-90 3 90 NaN
10 3-91 3 91 3-90
11 3-92 3 92 3-91
12 4-86 4 86 NaN
I tried to do this with a merge but I got Nan's all the way down in the 2nd half of the merge. Here's my code for that:
df['last_year'] = df['yr'].apply(lambda x: x-1 if x > 0 else None)
df_test = df.merge(df, how='left',indicator=False,left_on=['id','yr'],right_on=['id','last_year'])
I know there's a better way to do this, but I'm not sure what it is. can you help?
You can using shift
df['New']=df.sort_values(['id','yr']).groupby('id').data.shift()
df
Out[793]:
data id yr New
0 1-87 1 87 NaN
1 1-88 1 88 1-87
2 1-89 1 89 1-88
3 2-54 2 54 2-53
4 2-55 2 55 2-54
5 2-53 2 53 NaN
6 3-87 3 87 NaN
7 4-87 4 87 4-86
8 5-89 5 89 NaN
9 3-90 3 90 3-87
10 3-91 3 91 3-90
11 3-92 3 92 3-91
12 4-86 4 86 NaN

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