Sunday, January 23, 2022

[FIXED] How do i count values in multiple columns based on multiple criteria and create a new column row wise?

January 23, 2022 dataframe, pandas, python No comments

Issue

index	key	M1	M2	M3	M4	M5	M6	M7	M8	M9	M10	Average	Count_G10	Count_L10
0	a	12	0	159	0	20	49	0	131	157	153	68.1	4	3
1	b	0	68	195	189	0	79	12	179	21	62	80.5	3	4
2	c	0	139	0	188	12	0	31	87	152	73	68.2	4	2
3	d	126	156	0	112	178	146	0	19	192	25	95.4	6	2
4	e	109	0	172	0	0	0	44	145	186	100	75.6	5	1
5	f	63	183	194	183	0	163	136	13	163	162	126	6	2
6	g	101	143	0	184	0	107	103	0	60	133	83.1	6	1
7	h	13	101	139	86	101	72	93	151	0	0	75.6	6	1
8	i	182	71	73	73	129	32	56	135	0	114	86.5	4	5
9	j	82	0	198	0	117	21	0	32	64	146	66	4	2
10	k	145	0	194	0	156	71	0	89	57	31	74.3	4	2

I would like to get the columns count_G10 and count_L10 where the logic for count_G10 is as follows: count of months(M1 to M10 columns) where value is >0 and ((value-average)/average) > 0.1

Similarly, count_L10 logic is: count of months(M1 to M10 columns) where value is >0 and ((value-average)/average) < -0.1

I have tried the following in Pandas:

Oct20_Nov21 = [202010,202011,202012,202101,202102,202103,202104,202105,202106,202107,202108,202109,202110,202111] 

new_df3['G10%'] = new_df3[Oct20_Nov21].applymap(lambda x : 1 if (((x-new_df3.avg_w_s)/new_df3.avg_w_s) > 0.1).any() else 0).values.sum(axis=1)

I get the following error:

The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Let me know what am I missing here. Thank you

Solution

Don't use applymap as it is both very slow and will try to perform operations on scalar values and Series leading to the error shown.

Instead it is best to perform the operations at the DataFrame level.

In the most verbose way we could do:

m_df = df.filter(like='M')

df['Count_G10'] = (
        m_df.gt(0) &
        m_df.sub(df['Average'], axis=0).div(df['Average'], axis=0).gt(0.1)
).sum(axis=1)

df['Count_L10'] = (
        m_df.gt(0) &
        m_df.sub(df['Average'], axis=0).div(df['Average'], axis=0).lt(-0.1)
).sum(axis=1)

key	M1	M2	M3	M4	M5	M6	M7	M8	M9	M10	Average	Count_G10	Count_L10
a	12	0	159	0	20	49	0	131	157	153	68.1	4	3
b	0	68	195	189	0	79	12	179	21	62	80.5	3	4
c	0	139	0	188	12	0	31	87	152	73	68.2	4	2
d	126	156	0	112	178	146	0	19	192	25	95.4	6	2
e	109	0	172	0	0	0	44	145	186	100	75.6	5	1
f	63	183	194	183	0	163	136	13	163	162	126	6	2
g	101	143	0	184	0	107	103	0	60	133	83.1	6	1
h	13	101	139	86	101	72	93	151	0	0	75.6	6	1
i	182	71	73	73	129	32	56	135	0	114	86.5	4	5
j	82	0	198	0	117	21	0	32	64	146	66	4	2
k	145	0	194	0	156	71	0	89	57	31	74.3	4	2

First filter to select only the M columns (There are many other ways to select only the desired columns which would also work):

m_df = df.filter(like='M')

     M1   M2   M3   M4   M5   M6   M7   M8   M9  M10
0    12    0  159    0   20   49    0  131  157  153
1     0   68  195  189    0   79   12  179   21   62
2     0  139    0  188   12    0   31   87  152   73
3   126  156    0  112  178  146    0   19  192   25
4   109    0  172    0    0    0   44  145  186  100
5    63  183  194  183    0  163  136   13  163  162
6   101  143    0  184    0  107  103    0   60  133
7    13  101  139   86  101   72   93  151    0    0
8   182   71   73   73  129   32   56  135    0  114
9    82    0  198    0  117   21    0   32   64  146
10  145    0  194    0  156   71    0   89   57   31

Then use the comparison operations gt and lt to do the value comparison operations.

Step 1: check is for values gt 0:

m_df.gt(0)

       M1     M2     M3     M4     M5     M6     M7     M8     M9    M10
0    True  False   True  False   True   True  False   True   True   True
1   False   True   True   True  False   True   True   True   True   True
2   False   True  False   True   True  False   True   True   True   True
3    True   True  False   True   True   True  False   True   True   True
4    True  False   True  False  False  False   True   True   True   True
5    True   True   True   True  False   True   True   True   True   True
6    True   True  False   True  False   True   True  False   True   True
7    True   True   True   True   True   True   True   True  False  False
8    True   True   True   True   True   True   True   True  False   True
9    True  False   True  False   True   True  False   True   True   True
10   True  False   True  False   True   True  False   True   True   True

Evaluate: ((value - average) / average). Both operations need to align on axis=0.

Step 2: Subtract

m_df.sub(df['Average'], axis=0)

      M1    M2     M3     M4     M5    M6    M7     M8     M9   M10
0  -56.1 -68.1   90.9  -68.1  -48.1 -19.1 -68.1   62.9   88.9  84.9
1  -80.5 -12.5  114.5  108.5  -80.5  -1.5 -68.5   98.5  -59.5 -18.5
2  -68.2  70.8  -68.2  119.8  -56.2 -68.2 -37.2   18.8   83.8   4.8
3   30.6  60.6  -95.4   16.6   82.6  50.6 -95.4  -76.4   96.6 -70.4
4   33.4 -75.6   96.4  -75.6  -75.6 -75.6 -31.6   69.4  110.4  24.4
5  -63.0  57.0   68.0   57.0 -126.0  37.0  10.0 -113.0   37.0  36.0
6   17.9  59.9  -83.1  100.9  -83.1  23.9  19.9  -83.1  -23.1  49.9
7  -62.6  25.4   63.4   10.4   25.4  -3.6  17.4   75.4  -75.6 -75.6
8   95.5 -15.5  -13.5  -13.5   42.5 -54.5 -30.5   48.5  -86.5  27.5
9   16.0 -66.0  132.0  -66.0   51.0 -45.0 -66.0  -34.0   -2.0  80.0
10  70.7 -74.3  119.7  -74.3   81.7  -3.3 -74.3   14.7  -17.3 -43.3

Step 3: Divide

m_df.sub(df['Average'], axis=0).div(df['Average'], axis=0)

          M1        M2        M3        M4        M5        M6        M7        M8        M9       M10
0  -0.823789 -1.000000  1.334802 -1.000000 -0.706314 -0.280470 -1.000000  0.923642  1.305433  1.246696
1  -1.000000 -0.155280  1.422360  1.347826 -1.000000 -0.018634 -0.850932  1.223602 -0.739130 -0.229814
2  -1.000000  1.038123 -1.000000  1.756598 -0.824047 -1.000000 -0.545455  0.275660  1.228739  0.070381
3   0.320755  0.635220 -1.000000  0.174004  0.865828  0.530398 -1.000000 -0.800839  1.012579 -0.737945
4   0.441799 -1.000000  1.275132 -1.000000 -1.000000 -1.000000 -0.417989  0.917989  1.460317  0.322751
5  -0.500000  0.452381  0.539683  0.452381 -1.000000  0.293651  0.079365 -0.896825  0.293651  0.285714
6   0.215403  0.720818 -1.000000  1.214200 -1.000000  0.287605  0.239471 -1.000000 -0.277978  0.600481
7  -0.828042  0.335979  0.838624  0.137566  0.335979 -0.047619  0.230159  0.997354 -1.000000 -1.000000
8   1.104046 -0.179191 -0.156069 -0.156069  0.491329 -0.630058 -0.352601  0.560694 -1.000000  0.317919
9   0.242424 -1.000000  2.000000 -1.000000  0.772727 -0.681818 -1.000000 -0.515152 -0.030303  1.212121
10  0.951548 -1.000000  1.611036 -1.000000  1.099596 -0.044415 -1.000000  0.197847 -0.232840 -0.582773

Step 4: Compare (gt or lt depending)

m_df.sub(df['Average'], axis=0).div(df['Average'], axis=0).gt(0.1)

       M1     M2     M3     M4     M5     M6     M7     M8     M9    M10
0   False  False   True  False  False  False  False   True   True   True
1   False  False   True   True  False  False  False   True  False  False
2   False   True  False   True  False  False  False   True   True  False
3    True   True  False   True   True   True  False  False   True  False
4    True  False   True  False  False  False  False   True   True   True
5   False   True   True   True  False   True  False  False   True   True
6    True   True  False   True  False   True   True  False  False   True
7   False   True   True   True   True  False   True   True  False  False
8    True  False  False  False   True  False  False   True  False   True
9    True  False   True  False   True  False  False  False  False   True
10   True  False   True  False   True  False  False   True  False  False

Step 5: Find where both conditions are True with logical AND (&)

(
        m_df.gt(0) &
        m_df.sub(df['Average'], axis=0).div(df['Average'], axis=0).gt(0.1)
)

       M1     M2     M3     M4     M5     M6     M7     M8     M9    M10
0   False  False   True  False  False  False  False   True   True   True
1   False  False   True   True  False  False  False   True  False  False
2   False   True  False   True  False  False  False   True   True  False
3    True   True  False   True   True   True  False  False   True  False
4    True  False   True  False  False  False  False   True   True   True
5   False   True   True   True  False   True  False  False   True   True
6    True   True  False   True  False   True   True  False  False   True
7   False   True   True   True   True  False   True   True  False  False
8    True  False  False  False   True  False  False   True  False   True
9    True  False   True  False   True  False  False  False  False   True
10   True  False   True  False   True  False  False   True  False  False

Step 6: Count the number of True values in each row with sum (True is 1 and False is 0 (the additive identity) which is why sum works to count number of True values)

(
        m_df.gt(0) &
        m_df.sub(df['Average'], axis=0).div(df['Average'], axis=0).gt(0.1)
).sum(axis=1)

0     4
1     3
2     4
3     6
4     5
5     6
6     6
7     6
8     4
9     4
10    4
dtype: int64

An almost identical process occurs for Count_L10 the only difference is checking .lt(-0.1) instead of .gt(0.1).

This operation can be greatly simplified by extracting and reusing common operations and refactoring the expression:

m_df = df.filter(like='M')
# Shared Condition
m = m_df.gt(0)
# Values
v = m_df.div(df['Average'], axis=0) - 1

df['Count_G10'] = (m & v.gt(0.1)).sum(axis=1)
df['Count_L10'] = (m & v.lt(-0.1)).sum(axis=1)

Both conditions use the check for values greater than 0 so we can keep that in a variable (m) to use multiple times. Both expressions compare against the same expression ((value - average) / average) we can also store this in a variable v.

The expression ((value - average) / average) can also be simplified to just ((value / average) - 1).

Since ((v - a) / a) = ((v/a) - (a / a)) = ((v/a) - 1)

This will reduce overall computation time, at the expense of some readability, but produces the same results:

key	M1	M2	M3	M4	M5	M6	M7	M8	M9	M10	Average	Count_G10	Count_L10
a	12	0	159	0	20	49	0	131	157	153	68.1	4	3
b	0	68	195	189	0	79	12	179	21	62	80.5	3	4
c	0	139	0	188	12	0	31	87	152	73	68.2	4	2
d	126	156	0	112	178	146	0	19	192	25	95.4	6	2
e	109	0	172	0	0	0	44	145	186	100	75.6	5	1
f	63	183	194	183	0	163	136	13	163	162	126	6	2
g	101	143	0	184	0	107	103	0	60	133	83.1	6	1
h	13	101	139	86	101	72	93	151	0	0	75.6	6	1
i	182	71	73	73	129	32	56	135	0	114	86.5	4	5
j	82	0	198	0	117	21	0	32	64	146	66	4	2
k	145	0	194	0	156	71	0	89	57	31	74.3	4	2

Setup used:

import pandas as pd

df = pd.DataFrame({
    'key': ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k'],
    'M1': [12, 0, 0, 126, 109, 63, 101, 13, 182, 82, 145],
    'M2': [0, 68, 139, 156, 0, 183, 143, 101, 71, 0, 0],
    'M3': [159, 195, 0, 0, 172, 194, 0, 139, 73, 198, 194],
    'M4': [0, 189, 188, 112, 0, 183, 184, 86, 73, 0, 0],
    'M5': [20, 0, 12, 178, 0, 0, 0, 101, 129, 117, 156],
    'M6': [49, 79, 0, 146, 0, 163, 107, 72, 32, 21, 71],
    'M7': [0, 12, 31, 0, 44, 136, 103, 93, 56, 0, 0],
    'M8': [131, 179, 87, 19, 145, 13, 0, 151, 135, 32, 89],
    'M9': [157, 21, 152, 192, 186, 163, 60, 0, 0, 64, 57],
    'M10': [153, 62, 73, 25, 100, 162, 133, 0, 114, 146, 31],
    'Average': [68.1, 80.5, 68.2, 95.4, 75.6, 126.0, 83.1, 75.6, 86.5, 66.0,
                74.3]
})

Answered By - Henry Ecker

This Answer collected from stackoverflow and tested by PythonFixing community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0

Sunday, January 23, 2022

[FIXED] How do i count values in multiple columns based on multiple criteria and create a new column row wise?

Issue

Solution

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