I'd like to slice my data frame, based on some group conditions, and get all the groups whose last record has a negative value.
A B C D
1 a a 1
1 a a 2
1 a a 3
2 a a 1
2 a a -1
3 a a -1
3 a a -2
3 a a -3
Suppose this is my data frame, group by column A.
I want to get all the groups with a negative last value in column D.
output:
A B C D
2 a a 1
2 a a -1
3 a a -1
3 a a -2
3 a a -3
Column B and C are not related to the filter. But I need all the rows in each group, not only the last rows.
How to do this?
Use groupby to group by column A and filter over its result to filter which groups you want to keep.
Full functional example:
import pandas as pd
df = pd.DataFrame({"A":[1, 1, 1, 2, 2, 3, 3, 3],
"B": ["a"] *8,
"C": ["a"] *8,
"D": [1, 2, 3, 1, -1, -1, -2 , -3]
})
df.groupby('A').filter(lambda df_group: df_group['D'].iloc[-1] < 0)
Output:
A B C D
3 2 a a 1
4 2 a a -1
5 3 a a -1
6 3 a a -2
7 3 a a -3
You can pass last method to groupby.transform to get the last items and use lt to get a boolean mask that returns True if the last group value is less than 0 and filter df:
out = df[df.groupby('A')['D'].transform('last').lt(0)]
Output:
A B C D
3 2 a a 1
4 2 a a -1
5 3 a a -1
6 3 a a -2
7 3 a a -3
Use:
string = """A B C D
1 a a 1
1 a a 2
1 a a 3
2 a a 1
2 a a -1
3 a a -1
3 a a -2
3 a a -3"""
import numpy as np
data = [x.split() for x in string.split('\n')]
import pandas as pd
df = pd.DataFrame(np.array(data[1:]), columns = data[0])
df['D'] = df['D'].astype(int)
#solution
temp = df.groupby('A')['D'].last()
df[df['A'].isin(temp[temp<0].index)]
Output:
I need to add a series with previous rows only if a condition matches in current cell. Here's the dataframe:
import pandas as pd
data = {'col1': [1, 2, 1, 0, 0, 0, 0, 3, 2, 2, 0, 0]}
df = pd.DataFrame(data, columns=['col1'])
df['continuous'] = df.col1
print(df)
I need to +1 a cell with previous sum if it's value > 0 else -1. So, result I'm expecting is;
col1 continuous
0 1 1//+1 as its non-zero
1 2 2//+1 as its non-zero
2 1 3//+1 as its non-zero
3 0 2//-1 as its zero
4 0 1
5 0 0
6 0 0// not to go less than 0
7 3 1
8 2 2
9 2 3
10 0 2
11 0 1
Case 2 : where I want instead of >0 , I need <-0.1
data = {'col1': [-0.097112634,-0.092674324,-0.089176841,-0.087302284,-0.087351866,-0.089226185,-0.092242213,-0.096446987,-0.101620036,-0.105940337,-0.109484752,-0.113515648,-0.117848816,-0.121133266,-0.123824577,-0.126030136,-0.126630895,-0.126015218,-0.124235003,-0.122715224,-0.121746573,-0.120794916,-0.120291174,-0.120323152,-0.12053229,-0.121491186,-0.122625851,-0.123819704,-0.125751858,-0.127676591,-0.129339428,-0.132342431,-0.137119556,-0.142040092,-0.14837848,-0.15439201,-0.159282645,-0.161271982,-0.162377701,-0.162838307,-0.163204393,-0.164095634,-0.165496071,-0.167224488,-0.167057078,-0.165706164,-0.163301617,-0.161423938,-0.158669389,-0.156508912,-0.15508329,-0.15365104,-0.151958972,-0.150317528,-0.149234892,-0.148259354,-0.14737422,-0.145958527,-0.144633388,-0.143120273,-0.14145652,-0.139930163,-0.138774126,-0.136710524,-0.134692221,-0.132534879,-0.129921444,-0.127974949,-0.128294058,-0.129241763,-0.132263506,-0.137828981,-0.145549768,-0.154244588,-0.163125109,-0.171814857,-0.179911465,-0.186223859,-0.190653162,-0.194761064,-0.197988536,-0.200500606,-0.20260121,-0.204797089,-0.208281065,-0.211846904,-0.215312626,-0.218696339,-0.221489975,-0.221375209,-0.220996031,-0.218558429,-0.215936558,-0.213933531,-0.21242896,-0.209682125,-0.208196607,-0.206243585,-0.202190476,-0.19913106,-0.19703291,-0.194244664,-0.189609518,-0.186600526,-0.18160171,-0.175875689,-0.170767095,-0.167453329,-0.163516985,-0.161168703,-0.158197984,-0.156378046,-0.154794499,-0.153236804,-0.15187487,-0.151623385,-0.150628282,-0.149039072,-0.14826268,-0.147535739,-0.145557646,-0.142223729,-0.139343068,-0.135355686,-0.13047743,-0.125999173,-0.12218752,-0.117021996,-0.111542982,-0.106409901,-0.101904095,-0.097910825,-0.094683375,-0.092079967,-0.088953862,-0.086268097,-0.082907394,-0.080723466,-0.078117426,-0.075431993,-0.072079536,-0.068962411,-0.064831759,-0.061257701,-0.05830671,-0.053889968,-0.048972414,-0.044763431,-0.042162829,-0.039328369,-0.038968862,-0.040450835,-0.041974942,-0.042161609,-0.04280523,-0.042702428,-0.042593856,-0.043166561,-0.043691795,-0.044093492,-0.043965231,-0.04263305,-0.040836102,-0.039605133,-0.037204273,-0.034368645,-0.032293737,-0.029037983,-0.025509509,-0.022704668,-0.021346266,-0.019881524,-0.018675734,-0.017509566,-0.017148129,-0.016671088,-0.016015011,-0.016241862,-0.016416445,-0.016548878,-0.016475455,-0.016405742,-0.015567737,-0.014190101,-0.012373151,-0.010370329,-0.008131459,-0.006729419,-0.005667607,-0.004883919,-0.004841328,-0.005403019,-0.005343759,-0.005377974,-0.00548823,-0.004889709,-0.003884973,-0.003149113,-0.002975268,-0.00283163,-0.00322658,-0.003546589,-0.004233582,-0.004448617,-0.004706967,-0.007400356,-0.010104064,-0.01230257,-0.014430498,-0.016499501,-0.015348355,-0.013974229,-0.012845464,-0.012688459,-0.012552231,-0.013719074,-0.014404172,-0.014611632,-0.013401283,-0.011807386,-0.007417753,-0.003321279,0.000363954,0.004908491,0.010151584,0.013223831,0.016746553,0.02106351,0.024571507,0.027588073,0.031313637,0.034419301,0.037016545,0.038172954,0.038237253,0.038094387,0.037783779,0.036482515,0.036080763,0.035476154,0.034107081,0.03237083,0.030934259,0.029317076,0.028236195,0.027850758,0.024612491,0.01964433,0.015153308,0.009684456,0.003336172]}
df = pd.DataFrame(data, columns=['col1'])
lim = float(-0.1)
s = df['col1'].lt(lim)
out = s.where(s, -1).cumsum()
df['sol'] = out - out.where((out < 0) & (~s)).ffill().fillna(0)
print(df)
The key problem here, to me, is to control the out not to go below zero. With that in mind, we can mask the output where it's negative and adjust accordingly:
# a little longer data for corner case
df = pd.DataFrame({'col1': [1, 2, 1, 0, 0, 0, 0, 3, 2, 2, 0, 0,0,0,0,2,3,4]})
s = df.col1.gt(0)
out = s.where(s,-1).cumsum()
df['continuous'] = out - out.where((out<0)&(~s)).ffill().fillna(0)
Output:
col1 continuous
0 1 1
1 2 2
2 1 3
3 0 2
4 0 1
5 0 0
6 0 0
7 3 1
8 2 2
9 2 3
10 0 2
11 0 1
12 0 0
13 0 0
14 0 0
15 2 1
16 3 2
17 4 3
You can do this using cumsum function on booleans:
Give me a +1 whenever col1 is not zero:
(df.col1 != 0 ).cumsum()
Give me a -1 whenever col1 is zero:
- (df.col1 == 0 ).cumsum()
Then just add them together!
df['continuous'] = (df.col1 != 0 ).cumsum() - (df.col1 == 0 ).cumsum()
However this does not resolve the dropping below zero criteria you mentioned
I have the following Pandas DataFrame in Python:
import numpy as np
import pandas as pd
df = pd.DataFrame(np.array([[1, 2, 3], [3, 2, 1], [2, 1, 1]]),
columns=['a', 'b', 'c'])
df
It looks as the following when you output it:
a b c
0 1 2 3
1 3 2 1
2 2 1 1
I need to add 3 new columns, as column "d", column "e", and column "f".
Values in each new column will be determined based on the values of column "b" and column "c".
In a given row:
If the value of column "b" is bigger than the value of column "c", columns [d, e, f] will have the values [1, 0, 0].
If the value of column "b" is equal to the value of column "c", columns [d, e, f] will have the values [0, 1, 0].
If the value of column "b" is smaller than the value of column "c", columns [d, e, f] will have the values [0, 0, 1].
After this operation, the DataFrame needs to look as the following:
a b c d e f
0 1 2 3 0 0 1 # Since b smaller than c
1 3 2 1 1 0 0 # Since b bigger than c
2 2 1 1 0 1 0 # Since b = c
My original DataFrame is much bigger than the one in this example.
Is there a good way of doing this in Python without looping through the DataFrame?
You can use np.where to create condition vector and use str.get_dummies to create dummies
df['vec'] = np.where(df.b>df.c, 'd', np.where(df.b == df.c, 'e', 'f'))
df = df.assign(**df['vec'].str.get_dummies()).drop('vec',1)
a b c d e f
0 1 2 3 0 0 1
1 3 2 1 1 0 0
2 2 1 1 0 1 0
Let us try np.sign with get_dummies, -1 is c<b, 0 is c=b, 1 is c>b
df=df.join(np.sign(df.eval('c-b')).map({-1:'d',0:'e',1:'f'}).astype(str).str.get_dummies())
df
Out[29]:
a b c d e f
0 1 2 3 0 0 1
1 3 2 1 1 0 0
2 2 1 1 0 1 0
You simply harness the Boolean conditions you've already specified.
df["d"] = np.where(df.b > df.c, 1, 0)
df["e"] = np.where(df.b == df.c, 1, 0)
df["f"] = np.where(df.b < df.c, 1, 0)
Can I insert a column at a specific column index in pandas?
import pandas as pd
df = pd.DataFrame({'l':['a','b','c','d'], 'v':[1,2,1,2]})
df['n'] = 0
This will put column n as the last column of df, but isn't there a way to tell df to put n at the beginning?
see docs: http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.insert.html
using loc = 0 will insert at the beginning
df.insert(loc, column, value)
df = pd.DataFrame({'B': [1, 2, 3], 'C': [4, 5, 6]})
df
Out:
B C
0 1 4
1 2 5
2 3 6
idx = 0
new_col = [7, 8, 9] # can be a list, a Series, an array or a scalar
df.insert(loc=idx, column='A', value=new_col)
df
Out:
A B C
0 7 1 4
1 8 2 5
2 9 3 6
If you want a single value for all rows:
df.insert(0,'name_of_column','')
df['name_of_column'] = value
Edit:
You can also:
df.insert(0,'name_of_column',value)
df.insert(loc, column_name, value)
This will work if there is no other column with the same name. If a column, with your provided name already exists in the dataframe, it will raise a ValueError.
You can pass an optional parameter allow_duplicates with True value to create a new column with already existing column name.
Here is an example:
>>> df = pd.DataFrame({'b': [1, 2], 'c': [3,4]})
>>> df
b c
0 1 3
1 2 4
>>> df.insert(0, 'a', -1)
>>> df
a b c
0 -1 1 3
1 -1 2 4
>>> df.insert(0, 'a', -2)
Traceback (most recent call last):
File "", line 1, in
File "C:\Python39\lib\site-packages\pandas\core\frame.py", line 3760, in insert
self._mgr.insert(loc, column, value, allow_duplicates=allow_duplicates)
File "C:\Python39\lib\site-packages\pandas\core\internals\managers.py", line 1191, in insert
raise ValueError(f"cannot insert {item}, already exists")
ValueError: cannot insert a, already exists
>>> df.insert(0, 'a', -2, allow_duplicates = True)
>>> df
a a b c
0 -2 -1 1 3
1 -2 -1 2 4
You could try to extract columns as list, massage this as you want, and reindex your dataframe:
>>> cols = df.columns.tolist()
>>> cols = [cols[-1]]+cols[:-1] # or whatever change you need
>>> df.reindex(columns=cols)
n l v
0 0 a 1
1 0 b 2
2 0 c 1
3 0 d 2
EDIT: this can be done in one line ; however, this looks a bit ugly. Maybe some cleaner proposal may come...
>>> df.reindex(columns=['n']+df.columns[:-1].tolist())
n l v
0 0 a 1
1 0 b 2
2 0 c 1
3 0 d 2
Here is a very simple answer to this(only one line).
You can do that after you added the 'n' column into your df as follows.
import pandas as pd
df = pd.DataFrame({'l':['a','b','c','d'], 'v':[1,2,1,2]})
df['n'] = 0
df
l v n
0 a 1 0
1 b 2 0
2 c 1 0
3 d 2 0
# here you can add the below code and it should work.
df = df[list('nlv')]
df
n l v
0 0 a 1
1 0 b 2
2 0 c 1
3 0 d 2
However, if you have words in your columns names instead of letters. It should include two brackets around your column names.
import pandas as pd
df = pd.DataFrame({'Upper':['a','b','c','d'], 'Lower':[1,2,1,2]})
df['Net'] = 0
df['Mid'] = 2
df['Zsore'] = 2
df
Upper Lower Net Mid Zsore
0 a 1 0 2 2
1 b 2 0 2 2
2 c 1 0 2 2
3 d 2 0 2 2
# here you can add below line and it should work
df = df[list(('Mid','Upper', 'Lower', 'Net','Zsore'))]
df
Mid Upper Lower Net Zsore
0 2 a 1 0 2
1 2 b 2 0 2
2 2 c 1 0 2
3 2 d 2 0 2
A general 4-line routine
You can have the following 4-line routine whenever you want to create a new column and insert into a specific location loc.
df['new_column'] = ... #new column's definition
col = df.columns.tolist()
col.insert(loc, col.pop()) #loc is the column's index you want to insert into
df = df[col]
In your example, it is simple:
df['n'] = 0
col = df.columns.tolist()
col.insert(0, col.pop())
df = df[col]