I have a dataframe, grouped, with multiindex columns as below:
import pandas as pd
codes = ["one","two","three"];
colours = ["black", "white"];
textures = ["soft", "hard"];
N= 100 # length of the dataframe
df = pd.DataFrame({ 'id' : range(1,N+1),
'weeks_elapsed' : [random.choice(range(1,25)) for i in range(1,N+1)],
'code' : [random.choice(codes) for i in range(1,N+1)],
'colour': [random.choice(colours) for i in range(1,N+1)],
'texture': [random.choice(textures) for i in range(1,N+1)],
'size': [random.randint(1,100) for i in range(1,N+1)],
'scaled_size': [random.randint(100,1000) for i in range(1,N+1)]
}, columns= ['id', 'weeks_elapsed', 'code','colour', 'texture', 'size', 'scaled_size'])
grouped = df.groupby(['code', 'colour']).agg( {'size': [np.sum, np.average, np.size, pd.Series.idxmax],'scaled_size': [np.sum, np.average, np.size, pd.Series.idxmax]}).reset_index()
>> grouped
code colour size scaled_size
sum average size idxmax sum average size idxmax
0 one black 1031 60.647059 17 81 185.153944 10.891408 17 47
1 one white 481 37.000000 13 53 204.139249 15.703019 13 53
2 three black 822 48.352941 17 6 123.269405 7.251141 17 31
3 three white 1614 57.642857 28 50 285.638337 10.201369 28 37
4 two black 523 58.111111 9 85 80.908912 8.989879 9 88
5 two white 669 41.812500 16 78 82.098870 5.131179 16 78
[6 rows x 10 columns]
How can I flatten/merge the column index levels as: "Level1|Level2", e.g. size|sum, scaled_size|sum. etc? If this is not possible, is there a way to groupby() as I did above without creating multi-index columns?
There is potentially a better way, more pythonic way to flatten MultiIndex columns.
1. Use map and join with string column headers:
grouped.columns = grouped.columns.map('|'.join).str.strip('|')
print(grouped)
Output:
code colour size|sum size|average size|size size|idxmax \
0 one black 862 53.875000 16 14
1 one white 554 46.166667 12 18
2 three black 842 49.529412 17 90
3 three white 740 56.923077 13 97
4 two black 1541 61.640000 25 50
scaled_size|sum scaled_size|average scaled_size|size scaled_size|idxmax
0 6980 436.250000 16 77
1 6101 508.416667 12 13
2 7889 464.058824 17 64
3 6329 486.846154 13 73
4 12809 512.360000 25 23
2. Use map with format for column headers that have numeric data types.
grouped.columns = grouped.columns.map('{0[0]}|{0[1]}'.format)
Output:
code| colour| size|sum size|average size|size size|idxmax \
0 one black 734 52.428571 14 30
1 one white 1110 65.294118 17 88
2 three black 930 51.666667 18 3
3 three white 1140 51.818182 22 20
4 two black 656 38.588235 17 77
5 two white 704 58.666667 12 17
scaled_size|sum scaled_size|average scaled_size|size scaled_size|idxmax
0 8229 587.785714 14 57
1 8781 516.529412 17 73
2 10743 596.833333 18 21
3 10240 465.454545 22 26
4 9982 587.176471 17 16
5 6537 544.750000 12 49
3. Use list comprehension with f-string for Python 3.6+:
grouped.columns = [f'{i}|{j}' if j != '' else f'{i}' for i,j in grouped.columns]
Output:
code colour size|sum size|average size|size size|idxmax \
0 one black 1003 43.608696 23 76
1 one white 1255 59.761905 21 66
2 three black 777 45.705882 17 39
3 three white 630 52.500000 12 23
4 two black 823 54.866667 15 33
5 two white 491 40.916667 12 64
scaled_size|sum scaled_size|average scaled_size|size scaled_size|idxmax
0 12532 544.869565 23 27
1 13223 629.666667 21 13
2 8615 506.764706 17 92
3 6101 508.416667 12 43
4 7661 510.733333 15 42
5 6143 511.916667 12 49
you could always change the columns:
grouped.columns = ['%s%s' % (a, '|%s' % b if b else '') for a, b in grouped.columns]
Based on Scott Boston's answer,
little update(it will be work for 2 or more levels column):
temp.columns.map(lambda x: '|'.join([str(i) for i in x]))
Thank you, Boston!
Full credit to suraj's concise answer: https://stackoverflow.com/a/72616083/317797
df.columns = df.columns.map('_'.join)
Related
Suppose we have a dataset.
tmp = pd.DataFrame({'hi': [1,2,3,3,5,6,3,2,3,2,1],
'bye': [12,23,35,35,53,62,31,22,33,22,12],
'yes': [12,2,32,3,5,6,23,2,32,2,21],
'no': [1,92,93,3,95,6,33,2,33,22,1],
'maybe': [91,2,32,3,95,69,3,2,93,2,1]})
In python we can easily do tmp.groupby('hi').agg(total_bye = ('bye', sum)) to get the sum of bye for each group. However, if I want to reference multiple columns, what would be the fastest, most efficient and least amount of cleanly (easily readable) written code to do this in python? In particular, can I do this using df.groupby(my_cols).agg()? What are the fastest alternatives? I'm open (actually prefer) to using faster libraries than pandas such as dask or vaex.
For example, in R data.table we can do this pretty easily, and it's super fast
# In R, assume this object is a data.table
# In a single line, the below code groups by 'hi' and then creates my_new_col column based on if bye > 5 and yes <= 20, taking the sum of 'no' for each group.
tmp[, .(my_new_col = sum(ifelse(bye > 5 & yes < 20, no, 0))), by = 'hi']
# output 1
hi my_new_col
1: 1 1
2: 2 116
3: 3 3
4: 5 95
5: 6 6
# Similarly, we can even group by a rule instead of creating a new col to group by. See below
tmp[, .(my_new_col = sum(ifelse(bye > 5 & yes < 20, no, 0))), by = .(new_rule = ifelse(hi > 3, 1, 0))]
# output 2
new_rule my_new_col
1: 0 120
2: 1 101
# We can even apply multiple aggregate functions in parallel using data.table
agg_fns <- function(x) list(sum=sum(as.double(x), na.rm=T),
mean=mean(as.double(x), na.rm=T),
min=min(as.double(x), na.rm=T),
max=max(as.double(x), na.rm=T))
tmp[,
unlist(
list(N = .N, # add a N column (row count) to the summary
unlist(mclapply(.SD, agg_fns, mc.cores = 12), recursive = F)), # apply all agg_fns over all .SDcols
recursive = F),
.SDcols = !unique(c(names('hi'), as.character(unlist('hi'))))]
output 3:
N bye.sum bye.mean bye.min bye.max yes.sum yes.mean yes.min yes.max no.sum no.mean no.min
1: 11 340 30.90909 12 62 140 12.72727 2 32 381 34.63636 1
no.max maybe.sum maybe.mean maybe.min maybe.max
1: 95 393 35.72727 1 95
Do we have this same flexibility in python?
You can use agg on all wanted columns and add a prefix:
tmp.groupby('hi').agg('sum').add_prefix('total_')
output:
total_bye total_yes total_no total_maybe
hi
1 24 33 2 92
2 67 6 116 6
3 134 90 162 131
5 53 5 95 95
6 62 6 6 69
You can even combine columns and operations flexibly with a dictionary:
tmp.groupby('hi').agg(**{'%s_%s' % (label,c): (c, op)
for c in tmp.columns
for (label,op) in [('total', 'sum'), ('average', 'mean')]
})
output:
total_hi average_hi total_bye average_bye total_yes average_yes total_no average_no total_maybe average_maybe
hi
1 2 1 24 12.000000 33 16.5 2 1.000000 92 46.00
2 6 2 67 22.333333 6 2.0 116 38.666667 6 2.00
3 12 3 134 33.500000 90 22.5 162 40.500000 131 32.75
5 5 5 53 53.000000 5 5.0 95 95.000000 95 95.00
6 6 6 62 62.000000 6 6.0 6 6.000000 69 69.00
If my data looks like this
Index Country ted_Val1 sam_Val1 ... ted_Val10 sam_Val10
1 Australia 1 3 ... 20 5
2 Bambua 12 33 ... 15 56
3 Tambua 14 34 ... 10 58
df = pd.DataFrame([["Australia", 1, 3, 20, 5],
["Bambua", 12, 33, 15, 56],
["Tambua", 14, 34, 10, 58]
], columns=["Country", "ted_Val1", "sam_Val1", "ted_Val10", "sam_Val10"]
)
I'd like to subtract all 'val_' columns from all 'ted_' values using a list, creating a new column starting with 'dif_' such that:
Index Country ted_Val1 sam_Val1 diff_Val1 ... ted_Val10 sam_Val10 diff_val10
1 Australia 1 3 -2 ... 20 5 -15
2 Bambua 12 33 12 ... 15 56 -41
3 Tambua 14 34 14... 10 58 -48
so far I've got:
calc_vars = ['ted_Val1',
'sam_Val1',
'ted_Val10',
'sam_Val10']
for i in calc_vars:
df_diff['dif_' + str(i)] = df.['ted_' + str(i)] - df.['sam_' + str(i)]
but I'm getting errors, not sure where to go from here. As a warning this is dummy data and there can be several underscores in the names
IIUC you can use filter to choose the columns for subtraction (assuming your columns are properly sorted like your sample):
print (pd.concat([df, pd.DataFrame(df.filter(like="ted").to_numpy()-df.filter(like="sam").to_numpy(),
columns=["diff"+i.split("_")[-1] for i in df.columns if "ted_Val" in i])],1))
Country ted_Val1 sam_Val1 ted_Val10 sam_Val10 diff1 diff10
0 Australia 1 3 20 5 -2 15
1 Bambua 12 33 15 56 -21 -41
2 Tambua 14 34 10 58 -20 -48
try this,
calc_vars = ['ted_Val1', 'sam_Val1', 'ted_Val10', 'sam_Val10']
# extract even & odd values from calc_vars
# ['ted_Val1', 'ted_Val10'], ['sam_Val1', 'sam_Val10']
for ted, sam in zip(calc_vars[::2], calc_vars[1::2]):
df['diff_' + ted.split("_")[-1]] = df[ted] - df[sam]
Edit: if columns are not sorted,
ted_cols = sorted(df.filter(regex="ted_Val\d+"), key=lambda x : x.split("_")[-1])
sam_cols = sorted(df.filter(regex="sam_Val\d+"), key=lambda x : x.split("_")[-1])
for ted, sam in zip(ted_cols, sam_cols):
df['diff_' + ted.split("_")[-1]] = df[ted] - df[sam]
Country ted_Val1 sam_Val1 ted_Val10 sam_Val10 diff_Val1 diff_Val10
0 Australia 1 3 20 5 -2 15
1 Bambua 12 33 15 56 -21 -41
2 Tambua 14 34 10 58 -20 -48
I have a dataframe called df_location:
location = {'location_id': [1,2,3,4,5,6,7,8,9,10],
'temperature_value': [20,21,22,23,24,25,26,27,28,29],
'humidity_value':[60,61,62,63,64,65,66,67,68,69]}
df_location = pd.DataFrame(locations)
I have another dataframe called df_islands:
islands = {'island_id':[10,20,30,40,50,60],
'list_of_locations':[[1],[2,3],[4,5],[6,7,8],[9],[10]]}
df_islands = pd.DataFrame(islands)
Each island_id corresponds to one or more locations. As you can see, the locations are stored in a list.
What I'm trying to do is to search the list_of_locations for each unique location and merge it to df_location in a way where each island_id will correspond to a specific location.
Final dataframe should be the following:
merged = {'location_id': [1,2,3,4,5,6,7,8,9,10],
'temperature_value': [20,21,22,23,24,25,26,27,28,29],
'humidity_value':[60,61,62,63,64,65,66,67,68,69],
'island_id':[10,20,20,30,30,40,40,40,50,60]}
df_merged = pd.DataFrame(merged)
I don't know whether there is a method or function in python to do so. I would really appreciate it if someone can give me a solution to this problem.
The pandas method you're looking for to expand your df_islands dataframe is .explode(column_name). From there, rename your column to location_id and then join the dataframes using pd.merge(). It'll perform a SQL-like join method using the location_id as the key.
import pandas as pd
locations = {'location_id': [1,2,3,4,5,6,7,8,9,10],
'temperature_value': [20,21,22,23,24,25,26,27,28,29],
'humidity_value':[60,61,62,63,64,65,66,67,68,69]}
df_locations = pd.DataFrame(locations)
islands = {'island_id':[10,20,30,40,50,60],
'list_of_locations':[[1],[2,3],[4,5],[6,7,8],[9],[10]]}
df_islands = pd.DataFrame(islands)
df_islands = df_islands.explode(column='list_of_locations')
df_islands.columns = ['island_id', 'location_id']
pd.merge(df_locations, df_islands)
Out[]:
location_id temperature_value humidity_value island_id
0 1 20 60 10
1 2 21 61 20
2 3 22 62 20
3 4 23 63 30
4 5 24 64 30
5 6 25 65 40
6 7 26 66 40
7 8 27 67 40
8 9 28 68 50
9 10 29 69 60
The df.apply() method works here. It's a bit long-winded but it works:
df_location['island_id'] = df_location['location_id'].apply(
lambda x: [
df_islands['island_id'][i] \
for i in df_islands.index \
if x in df_islands['list_of_locations'][i]
# comment above line and use this instead if list is stored in a string
# if x in eval(df_islands['list_of_locations'][i])
][0]
)
First we select the final value we want if the if statement is True: df_islands['island_id'][i]
Then we loop over each column in df_islands by using df_islands.index
Then create the if statement which loops over all values in df_islands['list_of_locations'] and returns True if the value for df_location['location_id'] is in the list.
Finally, since we must contain this long statement in square brackets, it is a list. However, we know that there is only one value in the list so we can index it by using [0] at the end.
I hope this helps and happy for other editors to make the answer more legible!
print(df_location)
location_id temperature_value humidity_value island_id
0 1 20 60 10
1 2 21 61 20
2 3 22 62 20
3 4 23 63 30
4 5 24 64 30
5 6 25 65 40
6 7 26 66 40
7 8 27 67 40
8 9 28 68 50
9 10 29 69 60
I want to train a binary classification ML model with some data that I have; something like this:
df
y ch1_g1 ch2_g1 ch3_g1 ch1_g2 ch2_g2 ch3_g2
0 20 89 62 23 3 74
1 51 64 19 2 83 0
0 14 58 2 71 31 48
1 32 28 2 30 92 91
1 51 36 51 66 15 14
...
My target (y) depends on three characteristics from two groups, however I have an imbalance in my data, a count of values of my y target reveals that I have more zeros than ones in a ratio of about 2.68. I correct this by looping each row and randomly swapping values from group 1 to group 2 and viceversa, like this:
for index,row in df.iterrows():
choice = np.random.choice([0,1])
if row['y'] != choice:
df.loc[index, 'y'] = choice
for column in df.columns[1:]:
key = column.replace('g1', 'g2') if 'g1' in column else column.replace('g2', 'g1')
df.loc[index, column] = row[key]
Doing this reduce the ratio to no more than 1.3, so I was wondering if there is a more direct aproach using pandas methods.
¿Anyone have an idea how to accomplish this?
Whether or not swapping columns solves class unbalance aside, I would swap the whole data set, and randomly choose between the original and the swapped:
# Step 1: swap the columns
df1 = pd.concat((df.filter(regex='[^(_g1)]$'),
df.filter(regex='_g1$')),
axis=1)
# Step 2: rename the columns
df1.columns = df.columns
# random choice
np.random.seed(1)
is_original = np.random.choice([True,False], size=len(df))
# concat to make new dataset
pd.concat((df[is_original],df1[~is_original]))
Output:
y ch1_g1 ch2_g1 ch3_g1 ch1_g2 ch2_g2 ch3_g2
2 0 14 58 2 71 31 48
3 1 32 28 2 30 92 91
0 0 23 3 74 20 89 62
1 1 2 83 0 51 64 19
4 1 66 15 14 51 36 51
Notice that row with indexes 1,4 have g1 swap with g2.
It is the first time I use pandas and I do not really know how to deal with my problematic.
In fact I have 2 data frame:
import pandas
blast=pandas.read_table("blast")
cluster=pandas.read_table("cluster")
Here is an exemple of their contents:
>>> cluster
cluster_name seq_names
0 1 g1.t1_0035
1 1 g1.t1_0035_0042
2 119365 g1.t1_0042
3 90273 g1.t1_0042_0035
4 71567 g10.t1_0035
5 37976 g10.t1_0035_0042
6 22560 g10.t1_0042
7 90280 g10.t1_0042_0035
8 82698 g100.t1_0035
9 47392 g100.t1_0035_0042
10 28484 g100.t1_0042
11 22580 g100.t1_0042_0035
12 19474 g1000.t1_0035
13 5770 g1000.t1_0035_0042
14 29708 g1000.t1_0042
15 99776 g1000.t1_0042_0035
16 6283 g10000.t1_0035
17 39828 g10000.t1_0035_0042
18 25383 g10000.t1_0042
19 106614 g10000.t1_0042_0035
20 6285 g10001.t1_0035
21 13866 g10001.t1_0035_0042
22 121157 g10001.t1_0042
23 106615 g10001.t1_0042_0035
24 6286 g10002.t1_0035
25 113 g10002.t1_0035_0042
26 25397 g10002.t1_0042
27 106616 g10002.t1_0042_0035
28 4643 g10003.t1_0035
29 13868 g10003.t1_0035_0042
... ... ...
and
[78793 rows x 2 columns]
>>> blast
qseqid sseqid pident length mismatch \
0 g1.t1_0035_0042 g1.t1_0035_0042 100.0 286 0
1 g1.t1_0035_0042 g1.t1_0035 100.0 257 0
2 g1.t1_0035_0042 g9307.t1_0035 26.9 134 65
3 g2.t1_0035_0042 g2.t1_0035_0042 100.0 445 0
4 g2.t1_0035_0042 g2.t1_0035 95.8 451 3
5 g2.t1_0035_0042 g24520.t1_0042_0035 61.1 429 137
6 g2.t1_0035_0042 g9924.t1_0042 61.1 429 137
7 g2.t1_0035_0042 g1838.t1_0035 86.2 29 4
8 g3.t1_0035_0042 g3.t1_0035_0042 100.0 719 0
9 g3.t1_0035_0042 g3.t1_0035 84.7 753 62
10 g4.t1_0035_0042 g4.t1_0035_0042 100.0 242 0
11 g4.t1_0035_0042 g3.t1_0035 98.8 161 2
12 g5.t1_0035_0042 g5.t1_0035_0042 100.0 291 0
13 g5.t1_0035_0042 g3.t1_0035 93.1 291 0
14 g6.t1_0035_0042 g6.t1_0035_0042 100.0 152 0
15 g6.t1_0035_0042 g4.t1_0035 100.0 152 0
16 g7.t1_0035_0042 g7.t1_0035_0042 100.0 216 0
17 g7.t1_0035_0042 g5.t1_0035 98.1 160 3
18 g7.t1_0035_0042 g11143.t1_0042 46.5 230 99
19 g7.t1_0035_0042 g27537.t1_0042_0035 40.8 233 111
20 g3778.t1_0035_0042 g3778.t1_0035_0042 100.0 86 0
21 g3778.t1_0035_0042 g6174.t1_0035 98.0 51 1
22 g3778.t1_0035_0042 g20037.t1_0035_0042 100.0 50 0
23 g3778.t1_0035_0042 g37190.t1_0035 100.0 50 0
24 g3778.t1_0035_0042 g15112.t1_0042_0035 66.0 53 18
25 g3778.t1_0035_0042 g6061.t1_0042 66.0 53 18
26 g18109.t1_0035_0042 g18109.t1_0035_0042 100.0 86 0
27 g18109.t1_0035_0042 g33071.t1_0035 100.0 81 0
28 g18109.t1_0035_0042 g32810.t1_0035 96.4 83 3
29 g18109.t1_0035_0042 g17982.t1_0035_0042 98.6 72 1
... ... ... ... ... ...
if you stay focus on the cluster database, the first column correspond to the cluster ID and inside those clusters there are several sequences ID.
What I need to to is first to split all my cluster (in R it would be like: liste=split(x = data$V2, f = data$V1) )
And then, creat a function which displays the most similarity paires sequence within each cluster.
here is an exemple:
let's say I have two clusters (dataframe cluster):
cluster 1:
seq1
seq2
seq3
seq4
cluster 2:
seq5
seq6
seq7
...
On the blast dataframe there is on the 3th column the similarity between all sequences (all against all), so something like:
seq1 vs seq1 100
seq1 vs seq2 90
seq1 vs seq3 56
seq1 vs seq4 49
seq1 vs seq5 40
....
seq2 vs seq3 70
seq2 vs seq4 98
...
seq5 vs seq5 100
seq5 vs seq6 89
seq5 vs seq7 60
seq7 vs seq7 46
seq7 vs seq7 100
seq6 vs seq6 100
and what I need to get is :
cluster 1 (best paired sequences):
seq 1 vs seq 2
cluster2 (best paired sequences):
seq 5 vs seq6
...
So as you can see, I do not want to take into account the sequences paired by themselves
IF someone could give me some clues it would be fantastic.
Thank you all.
Firstly I assume that there are no Pairings in 'blast' with sequences from two different Clusters. In other words: in this solution the cluster-ID of a pairing will be evaluated by only one of the two sequence IDs.
Including cluster information and pairing information into one dataframe:
data = cluster.merge(blast, left_on='seq_names', right_on='qseqid')
Then the data should only contain pairings of different sequences:
data = data[data['qseqid']!=data['sseqid']]
To ignore pairings which have the same substrings in their seqid, the most readable way would be to add data columns with these data:
data['qspec'] = [seqid.split('_')[1] for seqid in data['qseqid'].values]
data['sspec'] = [seqid.split('_')[1] for seqid in data['sseqid'].values]
Now equal spec-values can be filtered the same way like it was done with equal seqids above:
data = data[data['qspec']!=data['sspec']]
In the end the data should be grouped by cluster-ID and within each group, the maximum of pident is of interest:
data_grpd = data.groupby('cluster_name')
result = data.loc[data_grpd['pident'].idxmax()]
The only drawback here - except the above mentioned assumption - is, that if there are several exactly equal max-values, only one of them would be taken into account.
Note: if you don't want the spec-columns to be of type string, you could easiliy turn them into integers on the fly by:
import numpy as np
data['qspec'] = [np.int(seqid.split('_')[1]) for seqid in data['qseqid'].values]
This merges the dataframes based first on sseqid, then on qseqid, and then returns results_df. Any with 100% match are filtered out. Let me know if this works. You can then order by cluster name.
blast = blast.loc[blast['pident'] != 100]
results_df = cluster.merge(blast, left_on='seq_names',right_on='sseqid')
results_df = results_df.append(cluster.merge(blast, left_on='seq_names',right_on='qseqid'))