python panda new column with order of values - python

I would like to make a new column with the order of the numbers in a list. I get 3,1,0,4,2,5 ( index of the lowest numbers ) but I would like to have a new column with 2,1,4,0,3,5 ( so if I look at a row i get the list and I get in what order this number comes in the total list. what am I doing wrong?
df = pd.DataFrame({'list': [4,3,6,1,5,9]})
df['order'] = df.sort_values(by='list').index
print(df)

What you're looking for is the rank:
import pandas as pd
df = pd.DataFrame({'list': [4,3,6,1,5,9]})
df['order'] = df['list'].rank().sub(1).astype(int)
Result:
list order
0 4 2
1 3 1
2 6 4
3 1 0
4 5 3
5 9 5
You can use the method parameter to control how to resolve ties.

Related

nunique compare two Pandas dataframe with duplicates and pivot them

My input:
df1 = pd.DataFrame({'frame':[ 1,1,1,2,3,0,1,2,2,2,3,4,4,5,5,5,8,9,9,10,],
'label':['GO','PL','ICV','CL','AO','AO','AO','ICV','PL','TI','PL','TI','PL','CL','CL','AO','TI','PL','ICV','ICV'],
'user': ['user1','user1','user1','user1','user1','user1','user1','user1','user1','user1','user1','user1','user1','user1','user1','user1','user1','user1','user1','user1']})
df2 = pd.DataFrame({'frame':[ 1, 1, 2, 3, 4,0,1,2,2,2,4,4,5,6,6,7,8,9,10,11],
'label':['ICV','GO', 'CL','TI','PI','AO','GO','ICV','TI','PL','ICV','TI','PL','CL','CL','CL','AO','AO','PL','ICV'],
'user': ['user2','user2','user2','user2','user2','user2','user2','user2','user2','user2','user2','user2','user2','user2','user2','user2','user2','user2','user2','user2']})
df_c = pd.concat([df1,df2])
I trying compare two df, frame by frame, and check if label in df1 existing in same frame in df2. And make some calucation with result (pivot for example)
That my code:
m_df = df1.merge(df2,on=['frame'],how='outer' )
m_df['cross']=m_df.apply(lambda row: 'Matched'
if row['label_x']==row['label_y']
else 'Mismatched', axis='columns')
pv_m_unq= pd.pivot_table(m_df,
columns='cross',
index='label_x',
values='frame',
aggfunc=pd.Series.nunique,fill_value=0,margins=True)
pv_mc = pd.pivot_table(m_df,
columns='cross',
index='label_x',
values='frame',
aggfunc=pd.Series.count,fill_value=0,margins=True)
but this creates a some problem:
first, I can calqulate "simple" total (column All) of matched and missmatched as descipted in picture, or its "duplicated" as AO in pv_m or wrong number as in CL in pv_m_unq
and second, I think merge method as I use int not clever way, because I get if frame+label repetead in df(its happens often), in merged df I get number row in df1 X number of rows in df2 for this specific frame+label
I think maybe there is a smarter way to compare df and pivot them?
You got the unexpected result on margin total because the margin is making use the same function passed to aggfunc (i.e. pd.Series.nunique in this case) for its calculation and the values of Matched and Mismatched in these 2 rows are both the same as 1 (hence only one unique value of 1). (You are currently getting the unique count of frame id's)
Probably, you can achieve more or less what you want by taking the count on them (including margin, Matched and Mismatched) instead of the unique count of frame id's, by using pd.Series.count instead in the last line of codes:
pv_m = pd.pivot_table(m_df,columns='cross',index='label_x',values='frame', aggfunc=pd.Series.count, margins=True, fill_value=0)
Result
cross Matched Mismatched All
label_x
AO 0 1 1
CL 1 0 1
GO 1 1 2
ICV 1 1 2
PL 0 2 2
All 3 5 8
Edit
If all you need is to have the All column being the sum of Matched and Mismatched, you can do it as follows:
Change your code of generating pv_m_unq without building margin:
pv_m_unq= pd.pivot_table(m_df,
columns='cross',
index='label_x',
values='frame',
aggfunc=pd.Series.nunique,fill_value=0)
Then, we create the column All as the sum of Matched and Mismatched for each row, as follows:
pv_m_unq['All'] = pv_m_unq['Matched'] + pv_m_unq['Mismatched']
Finally, create the row All as the sum of Matched and Mismatched for each column and append it as the last row, as follows:
row_All = pd.Series({'Matched': pv_m_unq['Matched'].sum(),
'Mismatched': pv_m_unq['Mismatched'].sum(),
'All': pv_m_unq['All'].sum()},
name='All')
pv_m_unq = pv_m_unq.append(row_All)
Result:
print(pv_m_unq)
Matched Mismatched All
label_x
AO 1 3 4
CL 1 2 3
GO 1 1 2
ICV 2 4 6
PL 1 5 6
TI 2 3 5
All 8 18 26
You can use isin() function like this:
df3 =df1[df1.label.isin(df2.label)]

changing index of 1 row in pandas

I have the the below df build from a pivot of a larger df. In this table 'week' is the the index (dtype = object) and I need to show week 53 as the first row instead of the last
Can someone advice please? I tried reindex and custom sorting but can't find the way
Thanks!
here is the table
Since you can't insert the row and push others back directly, a clever trick you can use is create a new order:
# adds a new column, "new" with the original order
df['new'] = range(1, len(df) + 1)
# sets value that has index 53 with 0 on the new column
# note that this comparison requires you to match index type
# so if weeks are object, you should compare df.index == '53'
df.loc[df.index == 53, 'new'] = 0
# sorts values by the new column and drops it
df = df.sort_values("new").drop('new', axis=1)
Before:
numbers
weeks
1 181519.23
2 18507.58
3 11342.63
4 6064.06
53 4597.90
After:
numbers
weeks
53 4597.90
1 181519.23
2 18507.58
3 11342.63
4 6064.06
One way of doing this would be:
import pandas as pd
df = pd.DataFrame(range(10))
new_df = df.loc[[df.index[-1]]+list(df.index[:-1])].reset_index(drop=True)
output:
0
9 9
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
Alternate method:
new_df = pd.concat([df[df["Year week"]==52], df[~(df["Year week"]==52)]])

How can I rename NaN columns in python pandas?

Good day everyone! I had trouble putting a nested dictionary as separate columns. However, I fixed it using the concat and json.normalize function. But for some reason the code I used removed all the column names and returned NaN as values for the columns...
Does someone knows how to fix this?
Code I used:
import pandas as pd
c = ['photo.photo_replace', 'photo.photo_remove', 'photo.photo_add', 'photo.photo_effect', 'photo.photo_brightness',
'photo.background_color', 'photo.photo_resize', 'photo.photo_rotate', 'photo.photo_mirror', 'photo.photo_layer_rearrange',
'photo.photo_move', 'text.text_remove', 'text.text_add', 'text.text_edit', 'text.font_select', 'text.text_color', 'text.text_style',
'text.background_color', 'text.text_align', 'text.text_resize', 'text.text_rotate', 'text.text_move', 'text.text_layer_rearrange']
df_edit = pd.concat([json_normalize(x)[c] for x in df['editables']], ignore_index=True)
df.columns = df.columns.str.split('.').str[1]
Current problem:
Result I want:
df= pd.DataFrame({
'A':[1,2,3],
'B':[3,3,3]
})
print(df)
A B
0 1 3
1 2 3
2 3 3
c=['new_name1','new_name2']
df.columns=c
print(df)
new_name1 new_name2
0 1 3
1 2 3
2 3 3
remember , lenght of column names (c) should be equal to column amount

How to feed random numbers as indices to pandas data frame?

I'm trying to get a random sample from two pandas frames. If rows (random) 2,5,8 are selected in frame A, then the same 2,5,8 rows must be selected from frame B. I did it by first generating a random sample and now want to use this sample as indices for rows for frame. How can I do it? The code should look like
idx = list(random.sample(range(X_train.shape[0]),5000))
lgstc_reg[i].fit(X_train[idx,:], y_train[idx,:])
However, running the code gives an error.
Use iloc:
indexes = [2,5,8] # in your case this is the randomly generated list
A.iloc[indexes]
B.iloc[indexes]
An alternative consistent sampling methodology would be to set a random seed, and then sample:
random_seed = 42
A.sample(3, random_state=random_seed)
B.sample(3, random_state=random_seed)
The sampled DataFrames will have the same index.
Hope this helps!
>>> df1
value ID
0 3 2
1 4 2
2 7 8
3 8 8
4 11 8
>>> df2
value distance
0 3 0
1 4 0
2 7 1
3 8 0
4 11 0
I have two data frames. I want to select randoms of df1 along with corresponding rows of df2.
First I create a sample_index which a list of random rows of df using Pandas inbuilt function sample. Now use this index to location these rows in df1 and df2 with the help of another inbuilt funciton loc.
>>> selection_index = df1.sample(2).index
>>> selection_index
Int64Index([3, 1], dtype='int64')
>>> df1.loc[selection_index]
value ID
3 8 8
1 4 2
>>> df2.loc[selection_index]
value distance
3 8 0
1 4 0
>>>
In your case, this would become somewhat like
idx = X_train.sample(5000).index
lgstc_reg[i].fit(X_train.loc[idx], y_train.loc[idx])

How can I keep all columns in a dataframe, plus add groupby, and sum?

I have a data frame with 5 fields. I want to copy 2 fields from this into a new data frame. This works fine. df1 = df[['task_id','duration']]
Now in this df1, when I try to group by task_id and sum duration, the task_id field drops off.
Before (what I have now).
After (what I'm trying to achieve).
So, for instance, I'm trying this:
df1['total'] = df1.groupby(['task_id'])['duration'].sum()
The result is:
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
I don't know why I can't just sum the values in a column and group by unique IDs in another column. Basically, all I want to do is preserve the original two columns (['task_id', 'duration']), sum duration, and calculate a percentage of duration in a new column named pct. This seems like a very simple thing but I can't get anything working. How can I get this straightened out?
The code will take care of having the columns retained and getting the sum.
df[['task_id', 'duration']].groupby(['task_id', 'duration']).size().reset_index(name='counts')
Setup:
X = np.random.choice([0,1,2], 20)
Y = np.random.uniform(2,10,20)
df = pd.DataFrame({'task_id':X, 'duration':Y})
Calculate pct:
df = pd.merge(df, df.groupby('task_id').agg(sum).reset_index(), on='task_id')
df['pct'] = df['duration_x'].divide(df['duration_y'])*100
df.drop('duration_y', axis=1) # Drops sum duration, remove this line if you want to see it.
Result:
duration_x task_id pct
0 8.751517 0 58.017921
1 6.332645 0 41.982079
2 8.828693 1 9.865355
3 2.611285 1 2.917901
4 5.806709 1 6.488531
5 8.045490 1 8.990189
6 6.285593 1 7.023645
7 7.932952 1 8.864436
8 7.440938 1 8.314650
9 7.272948 1 8.126935
10 9.162262 1 10.238092
11 7.834692 1 8.754639
12 7.989057 1 8.927129
13 3.795571 1 4.241246
14 6.485703 1 7.247252
15 5.858985 2 21.396850
16 9.024650 2 32.957771
17 3.885288 2 14.188966
18 5.794491 2 21.161322
19 2.819049 2 10.295091
disclaimer: All data is randomly generated in setup, however, calculations are straightforward and should be correct for any case.
I finally got everything working in the following way.
# group by and sum durations
df1 = df1.groupby('task_id', as_index=False).agg({'duration': 'sum'})
list(df1)
# find each task_id as relative percentage of whole
df1['pct'] = df1['duration']/(df1['duration'].sum())
df1 = pd.DataFrame(df1)

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