We Keep Coding

Compare row wise elements of a single column. If there are 2 continuous L then select lowest from High column and ignore other. Conversly if 2 L

High D_HIGH D_HIGH_H
33 46.57 0 0L
0 69.93 42 42H
1 86.44 68 68H
34 56.58 83 83L
35 67.12 125 125L
2 117.91 158 158H
36 94.51 186 186L
3 120.45 245 245H
4 123.28 254 254H
37 83.20 286 286L
In column D_HIGH_H there is L & H at end.
If there are two continuous H then the one having highest value in High column has to be selected and other has to be ignored(deleted).
If there are two continuous L then the one having lowest value in High column has to be selected and other has to be ignored(deleted).
If the sequence is H,L,H,L then no changes to be made.
Output I want is as follows:
High D_HIGH D_HIGH_H
33 46.57 0 0L
1 86.44 68 68H
34 56.58 83 83L
2 117.91 158 158H
36 94.51 186 186L
4 123.28 254 254H
37 83.20 286 286L
I tried various options using list map but did not work out.Also tried with groupby but no logical conclusion.

Here's one way:
g = ((l := df['D_HIGH_H'].str[-1]) != l.shift()).cumsum()
def f(x):
if (x['D_HIGH_H'].str[-1] == 'H').any():
return x.nlargest(1, 'D_HIGH')
return x.nsmallest(1, 'D_HIGH')
df.groupby(g, as_index=False).apply(f)
Output:
High D_HIGH D_HIGH_H
0 33 46.57 0 0L
1 1 86.44 68 68H
2 34 56.58 83 83L
3 2 117.91 158 158H
4 36 94.51 186 186L
5 4 123.28 254 254H
6 37 83.20 286 286L

You can use extract to get the letter, then compute a custom group and groupby.apply with a function that depends on the letter:
# extract letter
s = df['D_HIGH_H'].str.extract('(\D)$', expand=False)
# group by successive letters
# get the idxmin/idxmax depending on the type of letter
keep = (df['High']
.groupby([s, s.ne(s.shift()).cumsum()], sort=False)
.apply(lambda x: x.idxmin() if x.name[0] == 'L' else x.idxmax())
.tolist()
)
out = df.loc[keep]
Output:
High D_HIGH D_HIGH_H
33 46.57 0 0L
1 86.44 68 68H
34 56.58 83 83L
2 117.91 158 158H
36 94.51 186 186L
4 123.28 254 254H
37 83.20 286 286L

Replace blank value in dataframe based on another column condition

I have many blanks in a merged data set and I want to fill them with a condition.
My current code looks like this
import pandas as pd
import csv
import numpy as np
pd.set_option('display.max_columns', 500)
# Read all files into pandas dataframes
Jan = pd.read_csv(r'C:\~\Documents\Jan.csv')
Feb = pd.read_csv(r'C:\~\Documents\Feb.csv')
Mar = pd.read_csv(r'C:\~\Documents\Mar.csv')
Jan=pd.DataFrame({'Department':['52','5','56','70','7'],'Item':['2515','254','818','','']})
Feb=pd.DataFrame({'Department':['52','56','765','7','40'],'Item':['2515','818','524','','']})
Mar=pd.DataFrame({'Department':['7','70','5','8','52'],'Item':['45','','818','','']})
all_df_list = [Jan, Feb, Mar]
appended_df = pd.concat(all_df_list)
df = appended_df
df.to_csv(r"C:\~\Documents\SallesDS.csv", index=False)
Data set:
df
Department Item
52 2515
5 254
56 818
70
7 50
52 2515
56 818
765 524
7
40
7 45
70
5 818
8
52
What I want is to fill the empty cells in Item with a correspondent values of the Department column.
So If Department is 52 and Item is empty it should be filled with 2515
Department 7 and Item is empty fill it with 45
and the result should look like this
df
Department Item
52 2515
5 254
56 818
70
7 50
52 2515
56 818
765 524
7 45
40
7 45
70
5 818
8
52 2515
I tried the following method but non of them worked.
1
df.loc[(df['Item'].isna()) & (df['Department'].str.contains(52)), 'Item'] = 2515
df.loc[(df['Item'].isna()) & (df['Department'].str.contains(7)), 'Item'] = 45
2
df["Item"] = df["Item"].fillna(df["Department"])
df = df.replace({"Item":{"52":"2515", "7":"45"}})
both ethir return error or do not work
Answer:
Hi I have used the below code and it worked
b = [52]
df.Item=np.where(df.Department.isin(b),df.Item.fillna(2515),df.Item)
a = [7]
df.Item=np.where(df.Department.isin(a),df.Item.fillna(45),df.Item)
Hope it helps someone who face the same issue

The following solution first creates a map of each department and it's maximum corresponding item (assuming there is one), and then matches that item to a department with a blank item. Note that in your data frame, the empty items are an empty string ("") and not NaN.
Create a map:
values = df.groupby('Department').max()
values['Item'] = values['Item'].apply(lambda x: np.nan if x == "" else x)
values = values.dropna().reset_index()
Department Item
0 5 818
1 52 2515
2 56 818
3 7 45
4 765 524
Then use df.apply():
df['Item'] = df.apply(lambda x: values[values['Department'] == x['Department']]['Item'].values if x['Item'] == "" else x['Item'], axis=1)
In this case, the new values will have brackets around them. They can be removed with str.replace():
df['Item'] = df['Item'].astype(str).str.replace(r'\[|\'|\'|\]', "", regex=True)
The result:
Department Item
0 52 2515
1 5 254
2 56 818
3 70
4 7 45
0 52 2515
1 56 818
2 765 524
3 7 45
4 40
0 7 45
1 70
2 5 818
3 8
4 52 2515

Hi I have used the below code and it worked
b = [52]
df.Item=np.where(df.Department.isin(b),df.Item.fillna(2515),df.Item)
a = [7]
df.Item=np.where(df.Department.isin(a),df.Item.fillna(45),df.Item)
Hope it helps someone who face the same issue

Using groupby() for a dataframe in pandas resulted Index Error

I have this dataframe:
x y z parameter
0 26 24 25 Age
1 35 37 36 Age
2 57 52 54.5 Age
3 160 164 162 Hgt
4 182 163 172.5 Hgt
5 175 167 171 Hgt
6 95 71 83 Wgt
7 110 68 89 Wgt
8 89 65 77 Wgt
I'm using pandas to get this final result:
x y parameter
0 160 164 Hgt
1 182 163 Hgt
2 175 167 Hgt
I'm using groupby() to extract and isolate rows based on same parameter Hgt from the original dataframe
First, I added a column to set it as an index:
df = df.insert(0,'index', [count for count in range(df.shape[0])], True)
And the dataframe came out like this:
index x y z parameter
0 0 26 24 25 Age
1 1 35 37 36 Age
2 2 57 52 54.5 Age
3 3 160 164 162 Hgt
4 4 182 163 172.5 Hgt
5 5 175 167 171 Hgt
6 6 95 71 83 Wgt
7 7 110 68 89 Wgt
8 8 89 65 77 Wgt
Then, I used the following code to group based on index and extract the columns I need:
df1 = df.groupby('index')[['x', 'y','parameter']]
And the output was:
x y parameter
0 26 24 Age
1 35 37 Age
2 57 52 Age
3 160 164 Hgt
4 182 163 Hgt
5 175 167 Hgt
6 95 71 Wgt
7 110 68 Wgt
8 89 65 Wgt
After that, I used the following code to isolate only Hgt values:
df2 = df1[df1['parameter'] == 'Hgt']
When I ran df2, I got an error saying:
IndexError: Column(s) ['x', 'y', 'parameter'] already selected
Am I missing something here? What to do to get the final result?

Because you asked what you did wrong, let me point to useless/bad code.
Without any judgement (this is just to help you improve future code), almost everything is incorrect. It feels like a succession of complicated ways to do useless things. Let me give some details:
df = df.insert(0,'index', [count for count in range(df.shape[0])], True)
This seems a very convoluted way to do df.reset_index(). Even [count for count in range(df.shape[0])] could be have been simplified by using range(df.shape[0]) directly.
But this step is not even needed for a groupby as you can group by index level:
df.groupby(level=0)
But... the groupby is useless anyways as you only have single membered groups.
Also, when you do:
df1 = df.groupby('index')[['x', 'y','parameter']]
df1 is not a dataframe but a DataFrameGroupBy object. Very useful to store in a variable when you know what you're doing, this is however causing the error in your case as you thought this was a DataFrame. You need to apply an aggregation or transformation method of the DataFrameGroupBy object to get back a DataFrame, which you didn't (likely because, as seen above, there isn't much interesting to do on single-membered groups).
So when you run:
df1[df1['parameter'] == 'Hgt']
again, all is wrong as df1['parameter'] is equivalent to df.groupby('index')[['x', 'y','parameter']]['parameter'] (the cause of the error as you select twice 'parameter'). Even if you removed this error, the equality comparison would give a single True/False as you still have your DataFrameGroupBy and not a DataFrame, and this would incorrectly try to subselect an inexistent column of the DataFrameGroupBy.
I hope it helped!

Do you really need groupby?
>>> df.loc[df['parameter'] == 'Hgt', ['x', 'y', 'parameter']].reset_index(drop=True)
x y parameter
0 160 164 Hgt
1 182 163 Hgt
2 175 167 Hgt

Pandas: How to (cleanly) unpivot two columns with same category?

I'm trying to unpivot two columns inside a pandas dataframe. The transformation I seek would be the inverse of this question.
We start with a dataset that looks like this:
import pandas as pd
import numpy as np
df_orig = pd.DataFrame(data=np.random.randint(255, size=(4,5)),
columns=['accuracy','time_a','time_b','memory_a', 'memory_b'])
df_orig
accuracy time_a time_b memory_a memory_b
0 6 118 170 102 239
1 241 9 166 159 162
2 164 70 76 228 121
3 228 121 135 128 92
I wish to unpivot both themwmory and time columns, obtaining this dataset in result:
df
accuracy memory category time
0 6 102 a 118
1 241 159 a 9
2 164 228 a 70
3 228 128 a 121
12 6 239 b 170
13 241 162 b 166
14 164 121 b 76
15 228 92 b 135
So far I have managed to get my desired output using df.melt() twice plus some extra commands:
df = df_orig.copy()
# Unpivot memory columns
df = df.melt(id_vars=['accuracy','time_a', 'time_b'],
value_vars=['memory_a', 'memory_b'],
value_name='memory',
var_name='mem_cat')
# Unpivot time columns
df = df.melt(id_vars=['accuracy','memory', 'mem_cat'],
value_vars=['time_a', 'time_b'],
value_name='time',
var_name='time_cat')
# Keep only the 'a'/'b' as categories
df.mem_cat = df.mem_cat.str[-1]
df.time_cat = df.time_cat.str[-1]
# Keeping only the colums whose categories match (DIRTY!)
df = df[df.mem_cat==df.time_cat]
# Removing the duplicated category column.
df = df.drop(columns='time_cat').rename(columns={"mem_cat":'category'})
Given how easy it was to solve the inverse question, I believe my code is way too complex. Can anyone do it better?

Use wide_to_long:
np.random.seed(123)
df_orig = pd.DataFrame(data=np.random.randint(255, size=(4,5)),
columns=['accuracy','time_a','time_b','memory_a', 'memory_b'])
df = (pd.wide_to_long(df_orig.reset_index(),
stubnames=['time','memory'],
i='index',
j='category',
sep='_',
suffix='\w+')
.reset_index(level=1)
.reset_index(drop=True)
.rename_axis(None))
print (df)
category accuracy time memory
0 a 254 109 66
1 a 98 230 83
2 a 123 57 225
3 a 113 126 73
4 b 254 126 220
5 b 98 17 106
6 b 123 214 96
7 b 113 47 32

Iterating over pandas rows to get minimum

Here is my dataframe:
Date cell tumor_size(mm)
25/10/2015 113 51
22/10/2015 222 50
22/10/2015 883 45
20/10/2015 334 35
19/10/2015 564 47
19/10/2015 123 56
22/10/2014 345 36
13/12/2013 456 44
What I want to do is compare the size of the tumors detected on the different days. Let's consider the cell 222 as an example; I want to compare its size to different cells but detected on earlier days e.g. I will not compare its size with cell 883, because they were detected on the same day. Or I will not compare it with cell 113, because it was detected later on.
As my dataset is too large, I have iterate over the rows. If I explain it in a non-pythonic way:
for the cell 222:
get_size_distance(absolute value):
(50 - 35 = 15), (50 - 47 = 3), (50 - 56 = 6), (50 - 36 = 14), (44 - 36 = 8)
get_minumum = 3, I got this value when I compared it with 564, so I will name it as a pait for the cell 222
Then do it for the cell 883
The resulting output should look like this:
Date cell tumor_size(mm) pair size_difference
25/10/2015 113 51 222 1
22/10/2015 222 50 123 6
22/10/2015 883 45 456 1
20/10/2015 334 35 345 1
19/10/2015 564 47 456 3
19/10/2015 123 56 456 12
22/10/2014 345 36 456 8
13/12/2013 456 44 NaN NaN
I will really appreciate your help

It's not pretty, but I believe it does the trick
a = pd.read_clipboard()
# Cut off last row since it was a faulty date. You can skip this.
df = a.copy().iloc[:-1]
# Convert to dates and order just in case (not really needed I guess).
df['Date'] = df.Date.apply(lambda x: datetime.strptime(x, '%d/%m/%Y'))
df.sort_values('Date', ascending=False)
# Rename column
df = df.rename(columns={"tumor_size(mm)": 'tumor_size'})
# These will be our lists of pairs and size differences.
pairs = []
diffs = []
# Loop over all unique dates
for date in df.Date.unique():
# Only take dates earlier then current date.
compare_df = df.loc[df.Date < date].copy()
# Loop over each cell for this date and find the minimum
for row in df.loc[df.Date == date].itertuples():
# If no cells earlier are available use nans.
if compare_df.empty:
pairs.append(float('nan'))
diffs.append(float('nan'))
# Take lowest absolute value and fill in otherwise
else:
compare_df['size_diff'] = abs(compare_df.tumor_size - row.tumor_size)
row_of_interest = compare_df.loc[compare_df.size_diff == compare_df.size_diff.min()]
pairs.append(row_of_interest.cell.values[0])
diffs.append(row_of_interest.size_diff.values[0])
df['pair'] = pairs
df['size_difference'] = diffs
returns:
Date cell tumor_size pair size_difference
0 2015-10-25 113 51 222.0 1.0
1 2015-10-22 222 50 564.0 3.0
2 2015-10-22 883 45 564.0 2.0
3 2015-10-20 334 35 345.0 1.0
4 2015-10-19 564 47 345.0 11.0
5 2015-10-19 123 56 345.0 20.0
6 2014-10-22 345 36 NaN NaN