Efficient way to create column referencing its own previous value - python

I am trying to generate a few columns in a dataframe with datetime index based on a rule which references their own previous values. I have tried a for loop on the length of df as per below but looking for a cleaner solution if possible?
Because what I want to do in the end is get the stats of generated columns (C,D,E in below example) over a large number of A,B....
import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randn(30, 2), columns=list('AB'))
reset_level = 0.5
df['diff'] = df['A'].diff()
df['C'], df['D'], df['E'] = [0.0, 0.0, 0.0]
for i in range(1,len(df)):
if abs(df.iloc[i-1]['C'] + df.iloc[i]['diff']) > (reset_level):
df.iat[i,3] = 0.000
df.iat[i,4] = (df.iloc[i-1]['C'] + df.iloc[i]['diff'])
else:
df.iat[i,3] = (df.iloc[i-1]['C'] + df.iloc[i]['diff'])
df.iat[i,4] = 0.000
df.iat[i,5] = 0.5 * df.iloc[i]['D'] * df.iloc[i]['D']
Edit : Adding expected output below
A B diff C D E
0 -0.352725 1.429037 NaN 0.000000 0.000000 0.000000
1 -1.024418 -0.644302 -0.671693 0.000000 -0.671693 0.225585
2 0.401065 0.419555 1.425483 0.000000 1.425483 1.016001
3 -1.302484 0.724320 -1.703549 0.000000 -1.703549 1.451039
4 0.427035 0.835221 1.729518 0.000000 1.729518 1.495617
5 0.158694 -0.416741 -0.268340 -0.268340 0.000000 0.000000
6 0.921985 -0.490635 0.763291 0.494951 0.000000 0.000000
7 -0.835297 -1.036580 -1.757282 0.000000 -1.262331 0.796740
8 0.752060 -0.279206 1.587356 0.000000 1.587356 1.259850
9 1.795306 -1.554886 1.043246 0.000000 1.043246 0.544181
10 -0.405100 -0.361454 -2.200406 0.000000 -2.200406 2.420893
11 -0.253629 -0.627245 0.151471 0.151471 0.000000 0.000000
12 -0.820573 -0.212886 -0.566944 -0.415473 0.000000 0.000000
13 0.473439 2.532487 1.294012 0.000000 0.878539 0.385916
14 -1.395435 1.016338 -1.868875 0.000000 -1.868875 1.746346
15 -0.244269 -0.337820 1.151166 0.000000 1.151166 0.662592
16 -2.084977 -1.262249 -1.840708 0.000000 -1.840708 1.694103
17 0.666323 -1.696245 2.751300 0.000000 2.751300 3.784825
18 0.235207 -0.513903 -0.431115 -0.431115 0.000000 0.000000
19 1.386456 -0.149153 1.151249 0.000000 0.720134 0.259296
20 0.093456 -0.298154 -1.293000 0.000000 -1.293000 0.835925
21 0.690499 -1.687416 0.597043 0.000000 0.597043 0.178230
22 1.287530 -1.390260 0.597031 0.000000 0.597031 0.178223
23 1.828138 -0.288829 0.540608 0.000000 0.540608 0.146128
24 0.209666 -0.903385 -1.618472 0.000000 -1.618472 1.309727
25 -1.010678 0.615569 -1.220344 0.000000 -1.220344 0.744619
26 -1.799800 1.536332 -0.789122 0.000000 -0.789122 0.311357
27 0.611096 -1.033066 2.410896 0.000000 2.410896 2.906209
28 -0.532675 -0.091541 -1.143770 0.000000 -1.143770 0.654105
29 2.468137 -1.046117 3.000811 0.000000 3.000811 4.502435

I converted your for loop using a numpy array to hold the conditions and then np.where to replace the values according to your condition:
Define condition array
condition = np.abs(df.C.shift() + df["diff"]) > reset_level
Replace the values according to condition
df.iloc[:, 3] = np.where(condition, np.zeros((df.shape[0])), (df['C'].shift() + df['diff']))
df.iloc[:, 4] = np.where(~condition, np.zeros((df.shape[0])), (df['C'].shift() + df['diff']))
df.iloc[:, 5] = 0.5 * df['D'] * df['D']
Output:
A B diff C D E
0 -0.432513 -0.259526 NaN NaN 0.000000 0.000000
1 -1.120872 -1.572850 -0.688360 0.000000 NaN NaN
2 -0.917555 -2.251316 0.203317 0.203317 0.000000 0.000000
3 -1.869781 -1.284524 -0.952225 0.000000 -0.748908 0.280432
4 -2.041950 -0.091837 -0.172169 -0.172169 0.000000 0.000000
5 -0.142499 0.207746 1.899451 0.000000 1.727282 1.491751
6 1.432833 0.085211 1.575332 0.000000 1.575332 1.240835
7 -2.500191 -0.009907 -3.933025 0.000000 -3.933025 7.734341
8 0.154460 -1.859954 2.654651 0.000000 2.654651 3.523587
9 -0.565057 -0.516736 -0.719517 0.000000 -0.719517 0.258853
10 0.329845 0.127978 0.894902 0.000000 0.894902 0.400425
11 -0.920558 1.254617 -1.250402 0.000000 -1.250402 0.781753
12 -1.396913 0.262378 -0.476355 -0.476355 0.000000 0.000000
13 0.117336 -0.439932 1.514249 0.000000 1.037894 0.538612
14 -0.227066 2.565831 -0.344402 -0.344402 0.000000 0.000000
15 0.077750 0.195277 0.304816 0.304816 0.000000 0.000000
16 1.470611 -0.357213 1.392861 0.000000 1.697677 1.441053
17 -0.553844 0.339270 -2.024455 0.000000 -2.024455 2.049209
18 -0.259603 0.212839 0.294242 0.294242 0.000000 0.000000
19 0.605961 0.279599 0.865564 0.000000 1.159805 0.672574
20 -0.326706 -0.774350 -0.932667 0.000000 -0.932667 0.434934
21 -0.927601 -2.360751 -0.600895 0.000000 -0.600895 0.180537
22 -0.372085 0.986228 0.555516 0.000000 0.555516 0.154299
23 -0.687731 -2.966817 -0.315647 -0.315647 0.000000 0.000000
24 -0.041028 -0.328898 0.646703 0.000000 0.331057 0.054799
25 0.099489 0.275983 0.140517 0.140517 0.000000 0.000000
26 0.468274 -0.287097 0.368785 0.368785 0.000000 0.000000
27 0.497417 -0.588481 0.029143 0.029143 0.000000 0.000000
28 0.603178 2.243163 0.105761 0.105761 0.000000 0.000000
29 -0.643283 -1.051491 -1.246461 0.000000 -1.140700 0.650598
Is this what you were looking for, you didn't provide expected output.
Documentation:
np.where

Try this one (but don't iterate over all rows - it will do the whole column at once for you):
df["C_prev"] = df["C"].shift(1)

import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.randn(30, 2), columns=list('AB'))
reset_level = 0.5
df['diff'] = df['A'].diff()
df['C'], df['D'], df['E'] = [0.0, 0.0, 0.0]
Then apply a function to each row:
def f(row):
if abs(df.loc[row.name - 1, 'C'] + row['diff']) > reset_level:
C = 0.0
D = df.loc[row.name - 1, 'C'] + row['diff']
else:
C = df.loc[row.name - 1, 'C'] + row['diff']
D = 0.0
E = 0.5 * row['D'] * row['D']
return(pd.Series([C, D, E]))
df.loc[1:, ['C', 'D', 'E']] = df[1:].apply(f, axis=1)

Related

Eliminate unigrams in char-level TF-IDF

I would like to extract unique combinations of letters within words using the scikit-learn TF-IDF vectorizer for an NLP problem. However, I'm not interested in individual letters, but letter combinations, so that, e.g. "the" should produce "th" and "he" but not "t", "h" or "e". My understanding is I should be able to use ngram_range. However, using ngram_range=(2,3) is still returning unigrams.
Example:
from sklearn.feature_extraction.text import TfidfVectorizer
examples = ['The cat on the mat',
'Fast and bulbous']
tfidf = TfidfVectorizer(max_features=None,
analyzer='char_wb',
ngram_range=(2, 3))
data=tfidf.fit_transform(examples)
print(pd.DataFrame(data=data.todense(),
index=examples,
columns = tfidf.get_feature_names_out()))
gives me the 2- and 3-gram results as expected but also unigrams (i.e. I don't want "a", "b", etc.):
a an b bu c \
The cat on the mat 0.000000 0.000000 0.000000 0.000000 0.139994
Fast and bulbous 0.181053 0.181053 0.181053 0.181053 0.000000
ca f fa m ma ... \
The cat on the mat 0.139994 0.000000 0.000000 0.139994 0.139994 ...
Fast and bulbous 0.000000 0.181053 0.181053 0.000000 0.000000 ...
s st st t th \
The cat on the mat 0.000000 0.000000 0.000000 0.199213 0.279987
Fast and bulbous 0.181053 0.181053 0.181053 0.128821 0.000000
the ul ulb us us
The cat on the mat 0.279987 0.000000 0.000000 0.000000 0.000000
Fast and bulbous 0.000000 0.181053 0.181053 0.181053 0.181053
[2 rows x 53 columns]
I would've expected this output with ngram_range=(1,3) but not ngram_range=(2,3).
Edit:
I just noticed that "a" is extracted from "Fast and bulbous", presumably as it occurs as " a", i.e. with a space before the "a", but not in "The cat on the mat" as the "a" in "cat" is surrounded by "c" and "t". Likewise, "u" is not extracted as there is no space surrounding it in either text.
It seems like TfidfVectorizer is extracting bigrams including spaces. Is there a way to turn this off? (I though using analyzer='char_wb' searched within words rather than across words).
I constructed a callable to pass to analyzer. It is stolen from based on the function from the source code of TfidfVectorizer used when analyzer is set to 'char_wb':
def char_wb_ngrams(text_document, ngram_range):
"""Callable for TfidfVectorizer analyzer, based on _char_wb_ngrams from TfidfVectorizer source code at
https://github.com/scikit-learn/scikit-learn/blob/f3f51f9b6/sklearn/feature_extraction/text.py"""
ngrams = []
min_n, max_n = ngram_range
for w in text_document.lower().split():
# This line in _char_wb_ngrams pads words with spaces and needs to be removed:
#w = " " + w + " "
w_len = len(w)
for n in range(min_n, max_n + 1):
offset = 0
ngrams.append(w[offset : offset + n])
while offset + n < w_len:
offset += 1
ngrams.append(w[offset : offset + n])
if offset == 0: # count a short word (w_len < n) only once
break
return ngrams
This works on the example data from above:
from functools import partial
tfidf_no_space = TfidfVectorizer(max_features=None,
analyzer=partial(char_wb_ngrams, ngram_range=(2,3)),
ngram_range=(2, 3))
data=tfidf_no_space.fit_transform(examples)
print(pd.DataFrame(data=data.todense(),
index=examples,
columns = tfidf_no_space.get_feature_names_out()))
which yields
an and as ast at \
The cat on the mat 0.000000 0.000000 0.000000 0.000000 0.436436
Fast and bulbous 0.229416 0.229416 0.229416 0.229416 0.000000
bo bou bu bul ca ... \
The cat on the mat 0.000000 0.000000 0.000000 0.000000 0.218218 ...
Fast and bulbous 0.229416 0.229416 0.229416 0.229416 0.000000 ...
nd on ou ous st \
The cat on the mat 0.000000 0.218218 0.000000 0.000000 0.000000
Fast and bulbous 0.229416 0.000000 0.229416 0.229416 0.229416
th the ul ulb us
The cat on the mat 0.436436 0.436436 0.000000 0.000000 0.000000
Fast and bulbous 0.000000 0.000000 0.229416 0.229416 0.229416
[2 rows x 28 columns]
I'm not sure this would work with punctuation, though. It would be good to have a version that strips punctuation and also doesn't require the call to partial (which fixes ngram_range in the function).

How to multiply two data frame column by column?

I have to similarity between columns within the one data frame. The expected result is the same as the correlation matrix output, but the calculation function is different(I wrote my self calc function). So the calc function should get calc_func(column1, column2). The idea is to get similarities between columns. row size is not important. as an Output I expect (937,937) matrix.
Sample data
0011 0012 0013 0014 0015 0019 0111 0112 0121 0122 0123 0125 0129 0161 0168 0172 0174 0175 0176 0221 0222 0223 0224 0230 0241 0242 0243 0249 0251 0252 0341 0342 0344 0345 0351 0352 0353 0361 0362 0363 0371 0372 0411 0412 0421 0422 0423 0430 0441 0449 0452 0453 0459 0461 0471 0472 0481 0482 0483 0484 0485 0541 0542 0544 0545 0546 0547 0548 0561 0564 0566 0567 0571 0572 0573 0574 0575 0576 0577 0579 0581 0583 \
Reporter ISO
AFG 0.149474 0.699753 0.000000e+00 0.000000 0.000000 6.084805 0.000000e+00 0.013655 0.123035 0.000000e+00 0.000000 0.011263 0.000000 0.000000e+00 0.000000 0.000000e+00 0.000000 0.000000 0.000000 0.000000 0.000000 0.040835 0.009775 0.000000 0.000000 0.000000 1.902343e-04 0.003110 0.000000 0.000000 9.900480e-04 0.000382 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.002613 0.002373 0.000184 0.000000 0.000000e+00 5.570409e-04 0.000000 0.001367 0.023009 1.074305 0.000000 4.309246e-04 2.267049 0.135528 6.845710 0.000172 4.785010e-02 5.620574e-04 0.015391 0.000000 0.000000 0.008071 0.000000 0.602458 56.772035 4.902713e+01 11.542497 0.175537 0.000000 8.314311 6.899700e-01 0.009341 0.000000e+00 0.118446 0.465433 0.634222 0.008141 4.406345e+01 1.806608e+02 1457.266474 37.572639 16.111153 0.278868 5.828552e-01
AGO 0.000233 0.000000 2.169950e-05 0.000436 0.000021 0.206904 1.850937e-05 0.001081 0.000054 4.163925e-04 0.000437 0.000348 0.000059 1.287730e-04 0.000289 9.425705e-04 0.000140 0.002698 0.000444 0.000116 0.002252 0.000614 0.000295 0.000481 0.000003 0.000008 0.000000e+00 0.000142 0.000742 0.002136 3.700936e-01 1.594887 0.024370 0.000002 0.039695 0.146148 0.000020 0.267286 0.866269 0.036852 0.000384 0.046401 4.496454e-06 0.000000e+00 0.000834 0.016216 0.001110 0.000000 0.000065 7.354149e-04 0.000061 0.004332 0.000039 0.239055 2.501597e-01 3.830196e-04 0.000546 0.008450 0.015806 0.001086 0.002724 0.009187 0.005919 3.321169e-04 0.002146 0.000693 0.001050 0.006553 2.621612e-03 0.017289 8.732829e-05 0.000309 0.000343 0.000303 0.053893 5.683467e-04 6.637084e-05 0.000005 0.000036 0.001527 0.000022 1.886017e-07
ALB 0.004472 0.093826 0.000000e+00 0.000000 0.000000 4.959089 5.096002e-02 0.000000 0.000000 2.111634e-03 0.000487 0.003162 17.984117 1.681137e-02 0.000287 3.287117e-02 0.001309 0.001702 0.000093 0.005981 0.000027 0.004139 0.007258 0.000442 0.000000 0.122049 0.000000e+00 0.028040 1.376963 0.109314 2.201071e+00 0.646953 0.427123 0.055488 37.156633 24.666195 0.000416 0.452249 0.423161 1.855032 9.630443 16.673592 2.321445e-03 5.822343e-03 0.264946 0.000000 0.001616 0.000000 0.067036 1.468721e-03 0.000867 0.000000 0.000000 0.051276 3.280251e-02 1.379145e-02 0.026767 0.000000 0.315969 0.634852 0.004309 0.343613 0.302088 2.262782e+01 4.408535 0.013666 1.185906 1.818876 1.082149e+00 0.031302 5.695562e-03 3.008238 1.286605 0.064267 0.004062 1.028946e+00 5.242426e-02 2.020501 0.595951 1.282575 0.059749 8.487325e-01
Using exactly the example in the docs for df.corr():
def histogram_intersection(a, b):
v = np.minimum(a, b).sum().round(decimals=1)
return v
df = pd.DataFrame([(.2, .3), (.0, .6), (.6, .0), (.2, .1)],
columns=['dogs', 'cats'])
df.corr(method=histogram_intersection)
# output:
dogs cats
dogs 1.0 0.3
cats 0.3 1.0
So just pass your function in as the method parameter:
df.corr(method=calc_func)

Iterating over two list of dataframes of different length and concatenating as dataframes them within a loop to perform a function

Say, I have two lists of data frames. The size of the list of data frames are,
len(list_df1)
20
list_df1
[ Gene_id TCGA-A1-A0SM-01A-11R-A084-07 \
0 ENSG00000000003_PC 2.609759e+05
1 ENSG00000000005_PC 2.046517e+02
2 ENSG00000000419_PC 6.300425e+05
3 ENSG00000000457_PC 7.922487e+04
4 ENSG00000000460_PC 4.759842e+04
... ... ...
999 ENSG00000067167_PC 1.661041e+06
1000 ENSG00000067177_PC 1.117150e+05
1001 ENSG00000067182_PC 8.435913e+05
1002 ENSG00000067191_PC 1.219213e+05
1003 ENSG00000067208_PC 8.953860e+04
TCGA-A2-A04W-01A-31R-A115-07 TCGA-A2-A04X-01A-21R-A034-07 \
0 3.557029e+05 112350.116669
1 3.668742e+02 678.138127
2 9.395870e+05 980843.179895
3 7.663316e+04 54672.074426
4 3.543807e+04 21895.818992
... ... ...
999 1.306383e+06 652885.437844
1000 1.706889e+05 65106.955798
1001 1.255371e+06 634635.692737
1002 4.121517e+04 35705.970261
1003 7.729493e+04 51846.101598
TCGA-A2-A0D1-01A-11R-A034-07 TCGA-A2-A0EQ-01A-11R-A034-07 \
0 4.379606e+05 2.218208e+05
1 2.502301e+03 1.585833e+03
2 5.090120e+05 6.800754e+05
3 7.601743e+04 6.207091e+04
4 4.759907e+04 4.022122e+04
... ... ...
And for the second list of data frames,
len(list_df2)
15
list_df2
[ Gene_id TCGA-A1-A0SM-01A-11R-A084-07 \
1580 ENSG00000082929_lncRNAs 283.066402
1598 ENSG00000083622_lncRNAs 0.000000
1995 ENSG00000093100_lncRNAs 761.296884
2127 ENSG00000099869_lncRNAs 4134.407284
2819 ENSG00000103472_lncRNAs 4081.669994
... ... ...
25447 ENSG00000224648_lncRNAs 0.000000
25448 ENSG00000224649_lncRNAs 0.000000
25450 ENSG00000224652_lncRNAs 13698.679519
25452 ENSG00000224655_lncRNAs 454.154778
25455 ENSG00000224658_lncRNAs 0.000000
TCGA-A2-A04W-01A-31R-A115-07 TCGA-A2-A04X-01A-21R-A034-07 \
1580 16745.726795 0.000000
1598 0.000000 0.000000
1995 1091.806755 504.529752
2127 2058.792103 0.000000
2819 4541.654924 12125.955565
... ... ...
25447 0.000000 0.000000
25448 1806.322490 0.000000
25450 20679.823343 8560.813731
25452 0.000000 0.000000
25455 0.000000 0.000000
TCGA-A2-A0D1-01A-11R-A034-07 TCGA-A2-A0EQ-01A-11R-A034-07 \
1580 79172.350419 365.577246
1598 0.000000 0.000000
1995 698.135025 294.962048
2127 1053.164305 0.000000
2819 4517.449536 9633.998928
... ... ...
25447 0.000000 0.000000
25448 0.000000 0.000000
25450 6887.145309 931.142046
25452 347.062760 293.268032
25455 1619.254228 0.000000
TCGA-A2-A0T1-01A-21R-A084-07 TCGA-A2-A3XV-01A-21R-A239-07 \
1580 1024.779160 29452.491771
1598 0.000000 0.000000
1995 551.221323 243.727431
2127 415.769586 551.508707
2819 10530.579870 2027.695619
... ... ...
Now, what I need is, to concatenate the data frames from two lists into smaller data frames and run a function for each data frame.
Say my function is to perform correlation analysis,
import pandas as pd
from scipy.stats import pearsonr
def correlation(df):
correlations = pd.DataFrame()
for PC in [column for column in df.columns if '_PC' in column]:
for lncRNA in [column for column in df.columns if '_lncRNAs' in column]:
correlations = correlations.append(pd.Series(pearsonr(df[PC],df[lncRNA]),index=['PCC', 'p-value'],name=PC + '_' +lncRNA))
And I need to use the above function for each concatenated data frames from the lists above.
For example, this is what I tried.
for dfs,dfs2 in zip(list_df1,list_df2):
#print(dfs.iloc[0:2,0:2],dfs2.iloc[0:2,0:2])
df = dfs
df2 = dfs2
DF = pd.concat([dfs,dfs2])
DF_T =DF.set_index("Gene_id").T
The DF_T looks like following,
DF_T.shape()
(75, 1073)
However, DF_T is a concatenation of only last data frames from the 2 lists (list_df1,list_df2) which are defined above. Hence, if I run the correlation function on the data frame DF_T, it will only perform a correlation analysis only on the last data frames from the lists. Whereas, my aim is to run it all the small concatenated data frames from the two lists.
Therefore, I am looking for a way in which I can extract all data frames from two lists separately. Then, concatenate them into different data frames. Finally, apply the correlation function on each concatenated data frames. All these are done in order improve the performance speed of the correlation function. My input is multiple data frames of big size.
Any help or suggestions are much appreciated!!
Using your own code, this should work:
def correlation(list_df1,list_df2):
correlations = pd.DataFrame()
for dfs,dfs2 in zip(list_df1,list_df2):
DF = pd.concat([dfs,dfs2])
df = DF.set_index("Gene_id").T
for PC in [column for column in df.columns if '_PC' in column]:
for lncRNA in [column for column in df.columns if '_lncRNAs' in column]:
correlations = correlations.append(pd.Series(pearsonr(df[PC],df[lncRNA]),index=['PCC', 'p-value'],name=PC + '_' +lncRNA))
return correlations

ValueError: ndarray is not contiguous

when I build a matrix using the last row of my dataframe:
x = w.iloc[-1, :]
a = np.mat(x).T
it goes:
ValueError: ndarray is not contiguous
`print the x shows(I have 61 columns in my dataframe):
print(x)
cdl2crows 0.000000
cdl3blackcrows 0.000000
cdl3inside 0.000000
cdl3linestrike 0.000000
cdl3outside 0.191465
cdl3starsinsouth 0.000000
cdl3whitesoldiers_x 0.000000
cdl3whitesoldiers_y 0.000000
cdladvanceblock 0.000000
cdlhighwave 0.233690
cdlhikkake 0.218209
cdlhikkakemod 0.000000
...
cdlidentical3crows 0.000000
cdlinneck 0.000000
cdlinvertedhammer 0.351235
cdlkicking 0.000000
cdlkickingbylength 0.000000
cdlladderbottom 0.002259
cdllongleggeddoji 0.629053
cdllongline 0.588480
cdlmarubozu 0.065362
cdlmatchinglow 0.032838
cdlmathold 0.000000
cdlmorningdojistar 0.000000
cdlmorningstar 0.327749
cdlonneck 0.000000
cdlpiercing 0.251690
cdlrickshawman 0.471466
cdlrisefall3methods 0.000000
Name: 2010-01-04, Length: 61, dtype: float64
how to solve it? so many thanks
np.mat expects array form of input.
refer to the doc
doc
So your code should be
x = w.iloc[-1, :].values
a = np.mat(x).T
.values will give numpy array format of dataframe values, so np.mat will work.
Use np.array instead of np.mat:
a = np.array(x).T

TfIDf Vectorizer weights

Hi I have a lemmatized text in the format as shown by lemma. I want to get TfIdf score for each word this is the function that I wrote:
import numpy as np
import pandas as pd
from sklearn.feature_extraction.text import TfidfVectorizer
lemma=["'Ah", 'yes', u'say', 'softly', 'Harry',
'Potter', 'Our', 'new', 'celebrity', 'You',
'learn', 'subtle', 'science', 'exact', 'art',
'potion-making', u'begin', 'He', u'speak', 'barely',
'whisper', 'caught', 'every', 'word', 'like',
'Professor', 'McGonagall', 'Snape', 'gift',
u'keep', 'class', 'silent', 'without', 'effort',
'As', 'little', 'foolish', 'wand-waving', 'many',
'hardly', 'believe', 'magic', 'I', 'dont', 'expect', 'really',
'understand', 'beauty']
def Tfidf_Vectorize(lemmas_name):
vect = TfidfVectorizer(stop_words='english',ngram_range=(1,2))
vect_transform = vect.fit_transform(lemmas_name)
# First approach of creating a dataframe of weight & feature names
vect_score = np.asarray(vect_transform.mean(axis=0)).ravel().tolist()
vect_array = pd.DataFrame({'term': vect.get_feature_names(), 'weight': vect_score})
vect_array.sort_values(by='weight',ascending=False,inplace=True)
# Second approach of getting the feature names
vect_fn = np.array(vect.get_feature_names())
sorted_tfidf_index = vect_transform.max(0).toarray()[0].argsort()
print('Largest Tfidf:\n{}\n'.format(vect_fn[sorted_tfidf_index[:-11:-1]]))
return vect_array
tf_dataframe=Tfidf_Vectorize(lemma)
print(tf_dataframe.iloc[:5,:])
The output I am getting by:
print('Largest Tfidf:\n{}\n'.format(vect_fn[sorted_tfidf_index[:-11:-1]]))
is
Largest Tfidf:
[u'yes' u'fools' u'fury' u'gale' u'ghosts' u'gift' u'glory' u'glow' u'good'
u'granger']
The result of tf_dataframe
term weight
261 snape 0.027875
238 say 0.022648
211 potter 0.013937
181 mind 0.010453
123 harry 0.010453
60 dark 0.006969
75 dumbledore 0.006969
311 voice 0.005226
125 head 0.005226
231 ron 0.005226
Shouldn't both approaches lead to the same result of top features? I just want to calculate the tfidf scores and get the top 5 features/weight. What am i doing wrong?
I am not sure what I am looking at here but I have the feeling that you're using TfidfVectorizer incorrectly. However, please correct me in case I got the wrong idea of what you're trying.
So.. what you need is a list of documents which you feed to fit_transform(). From that you can construct a matrix where, for example, each column represents a document and each row a word. One cell in that matrix is the tf-idf score of the word i in document j.
Here's an example:
import numpy as np
import pandas as pd
from sklearn.feature_extraction.text import TfidfVectorizer
documents = [
"This is a document.",
"This is another document with slightly more text.",
"Whereas this is yet another document with even more text than the other ones.",
"This document is awesome and also rather long.",
"The car he drove was red."
]
document_names = ['Doc {:d}'.format(i) for i in range(len(documents))]
def get_tfidf(docs, ngram_range=(1,1), index=None):
vect = TfidfVectorizer(stop_words='english', ngram_range=ngram_range)
tfidf = vect.fit_transform(documents).todense()
return pd.DataFrame(tfidf, columns=vect.get_feature_names(), index=index).T
print(get_tfidf(documents, ngram_range=(1,2), index=document_names))
Which will give you:
Doc 0 Doc 1 Doc 2 Doc 3 Doc 4
awesome 0.0 0.000000 0.000000 0.481270 0.000000
awesome long 0.0 0.000000 0.000000 0.481270 0.000000
car 0.0 0.000000 0.000000 0.000000 0.447214
car drove 0.0 0.000000 0.000000 0.000000 0.447214
document 1.0 0.282814 0.282814 0.271139 0.000000
document awesome 0.0 0.000000 0.000000 0.481270 0.000000
document slightly 0.0 0.501992 0.000000 0.000000 0.000000
document text 0.0 0.000000 0.501992 0.000000 0.000000
drove 0.0 0.000000 0.000000 0.000000 0.447214
drove red 0.0 0.000000 0.000000 0.000000 0.447214
long 0.0 0.000000 0.000000 0.481270 0.000000
ones 0.0 0.000000 0.501992 0.000000 0.000000
red 0.0 0.000000 0.000000 0.000000 0.447214
slightly 0.0 0.501992 0.000000 0.000000 0.000000
slightly text 0.0 0.501992 0.000000 0.000000 0.000000
text 0.0 0.405004 0.405004 0.000000 0.000000
text ones 0.0 0.000000 0.501992 0.000000 0.000000
The two methods you show to get to words and their respective scores calculate the mean over all documents and fetch the max score of each word respectively.
So let's do this and compare the two methods:
df = get_tfidf(documents, ngram_range=(1,2), index=index)
print(pd.DataFrame([df.mean(1), df.max(1)], index=['score_mean', 'score_max']).T)
We can see that the scores are of course different.
score_mean score_max
awesome 0.096254 0.481270
awesome long 0.096254 0.481270
car 0.089443 0.447214
car drove 0.089443 0.447214
document 0.367353 1.000000
document awesome 0.096254 0.481270
document slightly 0.100398 0.501992
document text 0.100398 0.501992
drove 0.089443 0.447214
drove red 0.089443 0.447214
long 0.096254 0.481270
ones 0.100398 0.501992
red 0.089443 0.447214
slightly 0.100398 0.501992
slightly text 0.100398 0.501992
text 0.162002 0.405004
text ones 0.100398 0.501992
Note:
You can convince yourself that this does the same as calling min/max on the TfidfVectorizer:
vect = TfidfVectorizer(stop_words='english', ngram_range=(1,2))
tfidf = vect.fit_transform(documents)
print(tfidf.max(0))
print(tfidf.mean(0))

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