I'm trying to modify multiple column values in pandas.Dataframes with different increments in each column so that the values in each column do not overlap with each other when graphed on a line graph.
Here's the end goal of what I want to do: link
Let's say I have this kind of Dataframe:
Col1 Col2 Col3
0 0.3 0.2
1 1.1 1.2
2 2.2 2.4
3 3 3.1
but with hundreds of columns and thousands of values.
When graphing this on a line-graph on excel or matplotlib, the values overlap with each other, so I would like to separate each column by adding the same values for each column like so:
Col1(+0) Col2(+10) Col3(+20)
0 10.3 20.2
1 11.1 21.2
2 12.2 22.4
3 13 23.1
By adding the same value to one column and increasing by an increment of 10 over each column, I am able to see each line without it overlapping in one graph.
I thought of using loops and iterations to automate this value-adding process, but I couldn't find any previous solutions on Stackoverflow that addresses how I could change the increment value (e.g. from adding 0 in Col1 in one loop, then adding 10 to Col2 in the next loop) between different columns, but not within the values in a column. To make things worse, I'm a beginner with no clue about programming or data manipulation.
Since the data is in a CSV format, I first used Pandas to read it and store in a Dataframe, and selected the columns that I wanted to edit:
import pandas as pd
#import CSV file
df = pd.read_csv ('data.csv')
#store csv data into dataframe
df1 = pd.DataFrame (data = df)
# Locate columns that I want to edit with df.loc
columns = df1.loc[:, ' C000':]
here is where I'm stuck:
# use iteration with increments to add numbers
n = 0
for values in columns:
values = n + 0
print (values)
But this for-loop only adds one increment value (in this case 0), and adds it to all columns, not just the first column. Not only that, but I don't know how to add the next increment value for the next column.
Any possible solutions would be greatly appreciated.
IIUC ,just use df.add() over axis=1 with a list made from the length of df.columns:
df1 = df.add(list(range(0,len(df.columns)*10))[::10],axis=1)
Or as #jezrael suggested, better:
df1=df.add(range(0,len(df.columns)*10, 10),axis=1)
print(df1)
Col1 Col2 Col3
0 0 10.3 20.2
1 1 11.1 21.2
2 2 12.2 22.4
3 3 13.0 23.1
Details :
list(range(0,len(df.columns)*10))[::10]
#[0, 10, 20]
I would recommend you to avoid looping over the data frame as it is inefficient but rather think of adding to matrixes.
e.g.
import numpy as np
import pandas as pd
# Create your example df
df = pd.DataFrame(data=np.random.randn(10,3))
# Create a Matrix of ones
x = np.ones(df.shape)
# Multiply each column with an incremented value * 10
x = x * 10*np.arange(1,df.shape[1]+1)
# Add the matrix to the data
df + x
Edit: In case you do not want to increment with 10, 20 ,30 but 0,10,20 use this instead
import numpy as np
import pandas as pd
# Create your example df
df = pd.DataFrame(data=np.random.randn(10,3))
# Create a Matrix of ones
x = np.ones(df.shape)
# THIS LINE CHANGED
# Obmit the 1 so there is only an end value -> default start is 0
# Adjust the length of the vector
x = x * 10*np.arange(df.shape[1])
# Add the matrix to the data
df + x
Related
I have a problem where I would like to count the number of times the current value has not changed in a dataframe over rolling periods.
For example:
df = pd.DataFrame({'col':list('aaaabbab')})
would somehow give output of
0
1
2
3
0
1
0
0
I have been trying something along the following
df['col'] = df['col'] == df['col'].shift(1)
df.rolling(window=3).sum().reset_index(drop=True, level=0)
I have added in the rolling as I will want to look at the full data set in terms of rolling periods but even without having it over rolling periods I can not quite figure out the logic.
I am not sure if I am missing something simple or this may not be possible using shift
You need to generate a grouper for the change in values. For this compare each value with the previous one and apply a cumsum. This gives you groups in the itertools.groupby style ([1, 1, 1, 1, 2, 2, 3, 4]), finally group and apply a cumcount.
df['count'] = (df.groupby(df['col'].ne(df['col'].shift()).cumsum())
.cumcount()
)
output:
col count
0 a 0
1 a 1
2 a 2
3 a 3
4 b 0
5 b 1
6 a 0
7 b 0
edit: for fun here is a solution using itertools (much faster):
from itertools import groupby, chain
df['count'] = list(chain(*(list(range(len(list(g))))
for _,g in groupby(df['col']))))
NB. this runs much faster (88 µs vs 707 µs on the provided example)
I can't comment so just to add some more to #mozway answer.
My goal was to count consecutives value for an entire huge dataframe effectively.
The pb I encounter is that by construction
np.nan == np.nan
will return False so you could have a whole column full of only NaN and yet the counter will be at 0.
A simple workaround would be to replace all NaN in your df by a value not already in it.
For instance in the case of a float dataset you could do
df.fillna('NA')
which will work but by changing the dtype of your columns to Object the following code will be much slower (20x on my set up).
I would rather advised something like :
all_values = list(np.unique(np.array(df)))
all_values = [a for a in all_values if a==a]
unik_val = min(all_values)-1
temp = df.fillna(unik_val).copy()
from itertools import groupby, chain
for col in temp.columns:
temp[col] = list(chain(*(list(range(len(list(g))))
for _,g in groupby(temp[col]))))
count_df
I'm creating a Pandas dataframe from an existing file and it ends up essentially like this.
import pandas as pd
import datetime
data = [[i, i+1] for i in range(14)]
index = pd.date_range(start=datetime.date(2019,1,1), end=datetime.date(2020,2,1), freq='MS')
columns = ['col1', 'col2']
df = pd.DataFrame(data, index, columns)
Notice that this doesn't go all the way up to the present -- often the file I'm pulling from is a month or two behind. What I then need to do is add on any missing months and fill them with the same value as the previous year.
So in this case I need to add another row that is
2020-03-01 2 3
It could be anywhere from 0-2 rows that need to be added to the end of the dataframe at a given point in time. What's the best way to do this?
Note: The data here is not real so please don't take advantage of the simple pattern of entries I gave above. It was just a quick way to fill two columns of a table as an example.
If I understand your problem, then the following should help you. This does assume that you always have data 12 months ago however. You can define a new DataFrame which includes the months up to the most recent date.
# First create the new index. Get the most recent date and add an offset.
start, end = df.index[-1] + pd.DateOffset(), pd.Timestamp.now()
index_new = pd.date_range(start, end, freq='MS')
Create your DataFrame
# Get the data from the previous year.
data = df.loc[index_new - pd.DateOffset(years=1)].values
df_new = pd.DataFrame(data, index = index_new, columns=df.columns)
which looks like
col1 col2
2020-03-01 2 3
then just use;
pd.concat([df, df_new], axis=0)
Which gives
col1 col2
2019-01-01 0 1
2019-02-01 1 2
2019-03-01 2 3
... ... ...
2020-02-01 13 14
2020-03-01 2 3
Note
This also works for cases where the number of months missing is greater than 1.
Edit
Slightly different variation
# Create series with missing months added.
# Get the corresponding data 12 months prior.
s = pd.date_range(df.index[0], pd.Timestamp.now(), freq='MS')
fill = df.loc[s[~s.isin(df.index)] - pd.DateOffset(years=1)]
# Reindex the original dataframe
df = df.reindex(s)
# Find the dates to fill and replace with lagged data
df.iloc[-1 * fill.shape[0]:] = fill.values
Is there an easy method in pandas to invoke groupby on a range of values increments? For instance given the example below can I bin and group column B with a 0.155 increment so that for example, the first couple of groups in column B are divided into ranges between '0 - 0.155, 0.155 - 0.31 ...`
import numpy as np
import pandas as pd
df=pd.DataFrame({'A':np.random.random(20),'B':np.random.random(20)})
A B
0 0.383493 0.250785
1 0.572949 0.139555
2 0.652391 0.401983
3 0.214145 0.696935
4 0.848551 0.516692
Alternatively I could first categorize the data by those increments into a new column and subsequently use groupby to determine any relevant statistics that may be applicable in column A?
You might be interested in pd.cut:
>>> df.groupby(pd.cut(df["B"], np.arange(0, 1.0+0.155, 0.155))).sum()
A B
B
(0, 0.155] 2.775458 0.246394
(0.155, 0.31] 1.123989 0.471618
(0.31, 0.465] 2.051814 1.882763
(0.465, 0.62] 2.277960 1.528492
(0.62, 0.775] 1.577419 2.810723
(0.775, 0.93] 0.535100 1.694955
(0.93, 1.085] NaN NaN
[7 rows x 2 columns]
Try this:
df = df.sort_values('B')
bins = np.arange(0, 1.0, 0.155)
ind = np.digitize(df['B'], bins)
print df.groupby(ind).head()
Of course you can use any function on the groups not just head.
so this is how I use the groupby function
df1=data
bins = [0,40,50,60,70,100]
group_names=['F','S','C','B','A']
df1['grade']=pd.cut(data['student_mark'],bins,labels=group_names)
df1
I am inputting multiple spreadsheets with multiple columns of data. For each spreadsheet, the maximum value of each column is found. Then, for each element in the column, the element is divided by the maximum value of that column. The output should be a value (between 0 and 1) for each element in the column in ascending order. This is appended to a list which should be added to the source spreadsheet as a column.
Currently, the nested loops are performing correctly apart from the final step, as far as I understand. Each column is added to the spreadsheet EXCEPT the values are for the final column of the source spreadsheet rather than values related to each individual column.
I have tried changing the indents to associate levels of the code with different parts (as I think this is the problem) and tried moving the appended column along in the dataframe, to no avail.
for i in distlist:
#listname = i[4:] + '_norm'
df2 = pd.read_excel(i,header=0,index_col=None, skip_blank_lines=True)
df3 = df2.dropna(axis=0, how='any')
cols = []
for column in df3:
cols.append(column)
for x in cols:
listname = x + ' norm'
maxval = df3[x].max()
print(maxval)
mylist = []
for j in df3[x]:
findNL = (j/maxval)
mylist.append(findNL)
df3[listname] = mylist
saveloc = 'E:/test/'
filename = i[:-18] + '_Normalised.xlsx'
df3.to_excel(saveloc+filename, index=False)
New columns are added to the output dataframe with bespoke headings relating to the field headers in the source spreadsheet and renamed according to (listname). The data in each one of these new columns is identical and relates to the final column in the spreadsheet. To me, it seems to be overwriting the values each time (as if looping through the entire spreadsheet, not outputting for each column), and adding it to the spreadsheet.
Any help would be much appreciated. I think it's something simple, but I haven't managed to work out what...
If I understand you correctly, you are overcomplicating things. You dont need a for loop for this. You can simplify your code:
# Make example dataframe, this is not provided
df = pd.DataFrame({'col1':[1, 2, 3, 4],
'col2':[5, 6, 7, 8]})
print(df)
col1 col2
0 1 5
1 2 6
2 3 7
3 4 8
Now we can use DataFrame.apply and use add_suffix to give the new columns _norm suffix and after that concat the columns to one final dataframe
df_conc = pd.concat([df, df.apply(lambda x: x/x.max()).add_suffix('_norm')],axis=1)
print(df_conc)
col1 col2 col1_norm col2_norm
0 1 5 0.25 0.625
1 2 6 0.50 0.750
2 3 7 0.75 0.875
3 4 8 1.00 1.000
Many thanks. I think I was just overcomplicating it. Incidentally, I think my code may do the same job, but because there is so little difference in the values, it wasn't notable.
Thanks for your help #Erfan
How can I find the row for which the value of a specific column is maximal?
df.max() will give me the maximal value for each column, I don't know how to get the corresponding row.
Use the pandas idxmax function. It's straightforward:
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].idxmax()
3
>>> df['B'].idxmax()
4
>>> df['C'].idxmax()
1
Alternatively you could also use numpy.argmax, such as numpy.argmax(df['A']) -- it provides the same thing, and appears at least as fast as idxmax in cursory observations.
idxmax() returns indices labels, not integers.
Example': if you have string values as your index labels, like rows 'a' through 'e', you might want to know that the max occurs in row 4 (not row 'd').
if you want the integer position of that label within the Index you have to get it manually (which can be tricky now that duplicate row labels are allowed).
HISTORICAL NOTES:
idxmax() used to be called argmax() prior to 0.11
argmax was deprecated prior to 1.0.0 and removed entirely in 1.0.0
back as of Pandas 0.16, argmax used to exist and perform the same function (though appeared to run more slowly than idxmax).
argmax function returned the integer position within the index of the row location of the maximum element.
pandas moved to using row labels instead of integer indices. Positional integer indices used to be very common, more common than labels, especially in applications where duplicate row labels are common.
For example, consider this toy DataFrame with a duplicate row label:
In [19]: dfrm
Out[19]:
A B C
a 0.143693 0.653810 0.586007
b 0.623582 0.312903 0.919076
c 0.165438 0.889809 0.000967
d 0.308245 0.787776 0.571195
e 0.870068 0.935626 0.606911
f 0.037602 0.855193 0.728495
g 0.605366 0.338105 0.696460
h 0.000000 0.090814 0.963927
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
In [20]: dfrm['A'].idxmax()
Out[20]: 'i'
In [21]: dfrm.iloc[dfrm['A'].idxmax()] # .ix instead of .iloc in older versions of pandas
Out[21]:
A B C
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
So here a naive use of idxmax is not sufficient, whereas the old form of argmax would correctly provide the positional location of the max row (in this case, position 9).
This is exactly one of those nasty kinds of bug-prone behaviors in dynamically typed languages that makes this sort of thing so unfortunate, and worth beating a dead horse over. If you are writing systems code and your system suddenly gets used on some data sets that are not cleaned properly before being joined, it's very easy to end up with duplicate row labels, especially string labels like a CUSIP or SEDOL identifier for financial assets. You can't easily use the type system to help you out, and you may not be able to enforce uniqueness on the index without running into unexpectedly missing data.
So you're left with hoping that your unit tests covered everything (they didn't, or more likely no one wrote any tests) -- otherwise (most likely) you're just left waiting to see if you happen to smack into this error at runtime, in which case you probably have to go drop many hours worth of work from the database you were outputting results to, bang your head against the wall in IPython trying to manually reproduce the problem, finally figuring out that it's because idxmax can only report the label of the max row, and then being disappointed that no standard function automatically gets the positions of the max row for you, writing a buggy implementation yourself, editing the code, and praying you don't run into the problem again.
You might also try idxmax:
In [5]: df = pandas.DataFrame(np.random.randn(10,3),columns=['A','B','C'])
In [6]: df
Out[6]:
A B C
0 2.001289 0.482561 1.579985
1 -0.991646 -0.387835 1.320236
2 0.143826 -1.096889 1.486508
3 -0.193056 -0.499020 1.536540
4 -2.083647 -3.074591 0.175772
5 -0.186138 -1.949731 0.287432
6 -0.480790 -1.771560 -0.930234
7 0.227383 -0.278253 2.102004
8 -0.002592 1.434192 -1.624915
9 0.404911 -2.167599 -0.452900
In [7]: df.idxmax()
Out[7]:
A 0
B 8
C 7
e.g.
In [8]: df.loc[df['A'].idxmax()]
Out[8]:
A 2.001289
B 0.482561
C 1.579985
Both above answers would only return one index if there are multiple rows that take the maximum value. If you want all the rows, there does not seem to have a function.
But it is not hard to do. Below is an example for Series; the same can be done for DataFrame:
In [1]: from pandas import Series, DataFrame
In [2]: s=Series([2,4,4,3],index=['a','b','c','d'])
In [3]: s.idxmax()
Out[3]: 'b'
In [4]: s[s==s.max()]
Out[4]:
b 4
c 4
dtype: int64
df.iloc[df['columnX'].argmax()]
argmax() would provide the index corresponding to the max value for the columnX. iloc can be used to get the row of the DataFrame df for this index.
A more compact and readable solution using query() is like this:
import pandas as pd
df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
print(df)
# find row with maximum A
df.query('A == A.max()')
It also returns a DataFrame instead of Series, which would be handy for some use cases.
Very simple: we have df as below and we want to print a row with max value in C:
A B C
x 1 4
y 2 10
z 5 9
In:
df.loc[df['C'] == df['C'].max()] # condition check
Out:
A B C
y 2 10
If you want the entire row instead of just the id, you can use df.nlargest and pass in how many 'top' rows you want and you can also pass in for which column/columns you want it for.
df.nlargest(2,['A'])
will give you the rows corresponding to the top 2 values of A.
use df.nsmallest for min values.
The direct ".argmax()" solution does not work for me.
The previous example provided by #ely
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].argmax()
3
>>> df['B'].argmax()
4
>>> df['C'].argmax()
1
returns the following message :
FutureWarning: 'argmax' is deprecated, use 'idxmax' instead. The behavior of 'argmax'
will be corrected to return the positional maximum in the future.
Use 'series.values.argmax' to get the position of the maximum now.
So that my solution is :
df['A'].values.argmax()
mx.iloc[0].idxmax()
This one line of code will give you how to find the maximum value from a row in dataframe, here mx is the dataframe and iloc[0] indicates the 0th index.
Considering this dataframe
[In]: df = pd.DataFrame(np.random.randn(4,3),columns=['A','B','C'])
[Out]:
A B C
0 -0.253233 0.226313 1.223688
1 0.472606 1.017674 1.520032
2 1.454875 1.066637 0.381890
3 -0.054181 0.234305 -0.557915
Assuming one want to know the rows where column "C" is max, the following will do the work
[In]: df[df['C']==df['C'].max()])
[Out]:
A B C
1 0.472606 1.017674 1.520032
The idmax of the DataFrame returns the label index of the row with the maximum value and the behavior of argmax depends on version of pandas (right now it returns a warning). If you want to use the positional index, you can do the following:
max_row = df['A'].values.argmax()
or
import numpy as np
max_row = np.argmax(df['A'].values)
Note that if you use np.argmax(df['A']) behaves the same as df['A'].argmax().
Use:
data.iloc[data['A'].idxmax()]
data['A'].idxmax() -finds max value location in terms of row
data.iloc() - returns the row
If there are ties in the maximum values, then idxmax returns the index of only the first max value. For example, in the following DataFrame:
A B C
0 1 0 1
1 0 0 1
2 0 0 0
3 0 1 1
4 1 0 0
idxmax returns
A 0
B 3
C 0
dtype: int64
Now, if we want all indices corresponding to max values, then we could use max + eq to create a boolean DataFrame, then use it on df.index to filter out indexes:
out = df.eq(df.max()).apply(lambda x: df.index[x].tolist())
Output:
A [0, 4]
B [3]
C [0, 1, 3]
dtype: object
what worked for me is:
df[df['colX'] == df['colX'].max()
You then get the row in your df with the maximum value of colX.
Then if you just want the index you can add .index at the end of the query.