Pandas: new column using data from multiple other file - python-3.x

I would like to add a new column in a pandas dataframe df, filled with data that are in multiple other files.
Say my df is like this:
Sample Pos
A 5602
A 3069483
B 51948
C 231
And I have three files A_depth-file.txt, B_depth-file.txt, C_depth-file.txt like this (showing A_depth-file.txt):
Pos Depth
1 31
2 33
3 31
... ...
5602 52
... ...
3069483 40
The desired output df would have a new column Depth as follows:
Sample Pos Depth
A 5602 52
A 3069483 40
B 51948 32
C 231 47
I have a method that works but it takes about 20 minutes to fill a df with 712 lines, searching files of ~4 million lines (=positions). Would anyone know a better/faster way to do this?
The code I am using now is:
import pandas as pd
from io import StringIO
with open("mydf.txt") as f:
next(f)
List=[]
for line in f:
df = pd.read_fwf(StringIO(line), header=None)
df.rename(columns = {df.columns[1]: "Pos"}, inplace=True)
f2basename = df.iloc[:, 0].values[0]
f2 = f2basename + "_depth-file.txt"
df2 = pd.read_csv(f2, sep='\t')
df = pd.merge(df, df2, on="Pos", how="left")
List.append(df)
df = pd.concat(List, sort=False)
with open("mydf.txt") as f: to open the file to which I wish to add data
next(f) to pass the header
List=[] to create a new empty array called List
for line in f: to go over mydf.txt line by line and reading them with df = pd.read_fwf(StringIO(line), header=None)
df.rename(columns = {df.columns[1]: "Pos"}, inplace=True) to rename lost header name for Pos column, used later when merging line with associated file f2
f2basename = df.iloc[:, 0].values[0] getting basename of associated file f2 based on 1st column of mydf.txt
f2 = f2basename + "_depth-file.txt"to get full associated file f2 name
df2 = pd.read_csv(f2, sep='\t') to read file f2
df = pd.merge(df, df2, on="Pos", how="left")to merge the two files on column Pos, essentially adding Depth column to mydf.txt
List.append(df)adding modified line to the array List
df = pd.concat(List, sort=False) to concatenate elements of the List array into a dataframe df
Additional NOTES
In reality, I may need to search not only three files but several hundreds.

I didn't test the execution time, but should be faster if you read your 'mydf.txt' file in a dataframe too using read_csv and then use groupby and groupby apply.
If you know in advance that you have 3 samples and 3 relative files storing the depth, you can make a dictionary to read and store the three respective dataframes in advance and use them when needed.
df = pd.read_csv('mydf.txt', sep='\s+')
files = {basename : pd.read_csv(basename + "_depth-file.txt", sep='\s+') for basename in ['A', 'B', 'C']}
res = df.groupby('Sample').apply(lambda x : pd.merge(x, files[x.name], on="Pos", how="left"))
The final res would look like:
Sample Pos Depth
Sample
A 0 A 5602 52.0
1 A 3069483 40.0
B 0 B 51948 NaN
C 0 C 231 NaN
There are NaN values because I am using the sample provided and I don't have files for B and C (I used a copy of A), so values are missing. Provided that your files contain a 'Depth' for each 'Pos' you should not get any NaN.
To get rid of the multiindex made by groupby you can do:
res.reset_index(drop=True, inplace=True)
and res becomes:
Sample Pos Depth
0 A 5602 52.0
1 A 3069483 40.0
2 B 51948 NaN
3 C 231 NaN
EDIT after comments
Since you have a lot of files, you can use the following solution: same idea, but it does not require to read all the files in advance. Each file will be read when needed.
def merging_depth(x):
td = pd.read_csv(x.name + "_depth-file.txt", sep='\s+')
return pd.merge(x, td, on="Pos", how="left")
res = df.groupby('Sample').apply(merging_depth)
The result is the same.

Related

Adding a row to existing dataframe [duplicate]

How do I create an empty DataFrame, then add rows, one by one?
I created an empty DataFrame:
df = pd.DataFrame(columns=('lib', 'qty1', 'qty2'))
Then I can add a new row at the end and fill a single field with:
df = df._set_value(index=len(df), col='qty1', value=10.0)
It works for only one field at a time. What is a better way to add new row to df?
You can use df.loc[i], where the row with index i will be what you specify it to be in the dataframe.
>>> import pandas as pd
>>> from numpy.random import randint
>>> df = pd.DataFrame(columns=['lib', 'qty1', 'qty2'])
>>> for i in range(5):
>>> df.loc[i] = ['name' + str(i)] + list(randint(10, size=2))
>>> df
lib qty1 qty2
0 name0 3 3
1 name1 2 4
2 name2 2 8
3 name3 2 1
4 name4 9 6
In case you can get all data for the data frame upfront, there is a much faster approach than appending to a data frame:
Create a list of dictionaries in which each dictionary corresponds to an input data row.
Create a data frame from this list.
I had a similar task for which appending to a data frame row by row took 30 min, and creating a data frame from a list of dictionaries completed within seconds.
rows_list = []
for row in input_rows:
dict1 = {}
# get input row in dictionary format
# key = col_name
dict1.update(blah..)
rows_list.append(dict1)
df = pd.DataFrame(rows_list)
In the case of adding a lot of rows to dataframe, I am interested in performance. So I tried the four most popular methods and checked their speed.
Performance
Using .append (NPE's answer)
Using .loc (fred's answer)
Using .loc with preallocating (FooBar's answer)
Using dict and create DataFrame in the end (ShikharDua's answer)
Runtime results (in seconds):
Approach
1000 rows
5000 rows
10 000 rows
.append
0.69
3.39
6.78
.loc without prealloc
0.74
3.90
8.35
.loc with prealloc
0.24
2.58
8.70
dict
0.012
0.046
0.084
So I use addition through the dictionary for myself.
Code:
import pandas as pd
import numpy as np
import time
del df1, df2, df3, df4
numOfRows = 1000
# append
startTime = time.perf_counter()
df1 = pd.DataFrame(np.random.randint(100, size=(5,5)), columns=['A', 'B', 'C', 'D', 'E'])
for i in range( 1,numOfRows-4):
df1 = df1.append( dict( (a,np.random.randint(100)) for a in ['A','B','C','D','E']), ignore_index=True)
print('Elapsed time: {:6.3f} seconds for {:d} rows'.format(time.perf_counter() - startTime, numOfRows))
print(df1.shape)
# .loc w/o prealloc
startTime = time.perf_counter()
df2 = pd.DataFrame(np.random.randint(100, size=(5,5)), columns=['A', 'B', 'C', 'D', 'E'])
for i in range( 1,numOfRows):
df2.loc[i] = np.random.randint(100, size=(1,5))[0]
print('Elapsed time: {:6.3f} seconds for {:d} rows'.format(time.perf_counter() - startTime, numOfRows))
print(df2.shape)
# .loc with prealloc
df3 = pd.DataFrame(index=np.arange(0, numOfRows), columns=['A', 'B', 'C', 'D', 'E'] )
startTime = time.perf_counter()
for i in range( 1,numOfRows):
df3.loc[i] = np.random.randint(100, size=(1,5))[0]
print('Elapsed time: {:6.3f} seconds for {:d} rows'.format(time.perf_counter() - startTime, numOfRows))
print(df3.shape)
# dict
startTime = time.perf_counter()
row_list = []
for i in range (0,5):
row_list.append(dict( (a,np.random.randint(100)) for a in ['A','B','C','D','E']))
for i in range( 1,numOfRows-4):
dict1 = dict( (a,np.random.randint(100)) for a in ['A','B','C','D','E'])
row_list.append(dict1)
df4 = pd.DataFrame(row_list, columns=['A','B','C','D','E'])
print('Elapsed time: {:6.3f} seconds for {:d} rows'.format(time.perf_counter() - startTime, numOfRows))
print(df4.shape)
P.S.: I believe my realization isn't perfect, and maybe there is some optimization that could be done.
You could use pandas.concat(). For details and examples, see Merge, join, and concatenate.
For example:
def append_row(df, row):
return pd.concat([
df,
pd.DataFrame([row], columns=row.index)]
).reset_index(drop=True)
df = pd.DataFrame(columns=('lib', 'qty1', 'qty2'))
new_row = pd.Series({'lib':'A', 'qty1':1, 'qty2': 2})
df = append_row(df, new_row)
NEVER grow a DataFrame!
Yes, people have already explained that you should NEVER grow a DataFrame, and that you should append your data to a list and convert it to a DataFrame once at the end. But do you understand why?
Here are the most important reasons, taken from my post here.
It is always cheaper/faster to append to a list and create a DataFrame in one go.
Lists take up less memory and are a much lighter data structure to work with, append, and remove.
dtypes are automatically inferred for your data. On the flip side, creating an empty frame of NaNs will automatically make them object, which is bad.
An index is automatically created for you, instead of you having to take care to assign the correct index to the row you are appending.
This is The Right Way™ to accumulate your data
data = []
for a, b, c in some_function_that_yields_data():
data.append([a, b, c])
df = pd.DataFrame(data, columns=['A', 'B', 'C'])
These options are horrible
append or concat inside a loop
append and concat aren't inherently bad in isolation. The
problem starts when you iteratively call them inside a loop - this
results in quadratic memory usage.
# Creates empty DataFrame and appends
df = pd.DataFrame(columns=['A', 'B', 'C'])
for a, b, c in some_function_that_yields_data():
df = df.append({'A': i, 'B': b, 'C': c}, ignore_index=True)
# This is equally bad:
# df = pd.concat(
# [df, pd.Series({'A': i, 'B': b, 'C': c})],
# ignore_index=True)
Empty DataFrame of NaNs
Never create a DataFrame of NaNs as the columns are initialized with
object (slow, un-vectorizable dtype).
# Creates DataFrame of NaNs and overwrites values.
df = pd.DataFrame(columns=['A', 'B', 'C'], index=range(5))
for a, b, c in some_function_that_yields_data():
df.loc[len(df)] = [a, b, c]
The Proof is in the Pudding
Timing these methods is the fastest way to see just how much they differ in terms of their memory and utility.
Benchmarking code for reference.
It's posts like this that remind me why I'm a part of this community. People understand the importance of teaching folks getting the right answer with the right code, not the right answer with wrong code. Now you might argue that it is not an issue to use loc or append if you're only adding a single row to your DataFrame. However, people often look to this question to add more than just one row - often the requirement is to iteratively add a row inside a loop using data that comes from a function (see related question). In that case it is important to understand that iteratively growing a DataFrame is not a good idea.
If you know the number of entries ex ante, you should preallocate the space by also providing the index (taking the data example from a different answer):
import pandas as pd
import numpy as np
# we know we're gonna have 5 rows of data
numberOfRows = 5
# create dataframe
df = pd.DataFrame(index=np.arange(0, numberOfRows), columns=('lib', 'qty1', 'qty2') )
# now fill it up row by row
for x in np.arange(0, numberOfRows):
#loc or iloc both work here since the index is natural numbers
df.loc[x] = [np.random.randint(-1,1) for n in range(3)]
In[23]: df
Out[23]:
lib qty1 qty2
0 -1 -1 -1
1 0 0 0
2 -1 0 -1
3 0 -1 0
4 -1 0 0
Speed comparison
In[30]: %timeit tryThis() # function wrapper for this answer
In[31]: %timeit tryOther() # function wrapper without index (see, for example, #fred)
1000 loops, best of 3: 1.23 ms per loop
100 loops, best of 3: 2.31 ms per loop
And - as from the comments - with a size of 6000, the speed difference becomes even larger:
Increasing the size of the array (12) and the number of rows (500) makes
the speed difference more striking: 313ms vs 2.29s
mycolumns = ['A', 'B']
df = pd.DataFrame(columns=mycolumns)
rows = [[1,2],[3,4],[5,6]]
for row in rows:
df.loc[len(df)] = row
You can append a single row as a dictionary using the ignore_index option.
>>> f = pandas.DataFrame(data = {'Animal':['cow','horse'], 'Color':['blue', 'red']})
>>> f
Animal Color
0 cow blue
1 horse red
>>> f.append({'Animal':'mouse', 'Color':'black'}, ignore_index=True)
Animal Color
0 cow blue
1 horse red
2 mouse black
For efficient appending, see How to add an extra row to a pandas dataframe and Setting With Enlargement.
Add rows through loc/ix on non existing key index data. For example:
In [1]: se = pd.Series([1,2,3])
In [2]: se
Out[2]:
0 1
1 2
2 3
dtype: int64
In [3]: se[5] = 5.
In [4]: se
Out[4]:
0 1.0
1 2.0
2 3.0
5 5.0
dtype: float64
Or:
In [1]: dfi = pd.DataFrame(np.arange(6).reshape(3,2),
.....: columns=['A','B'])
.....:
In [2]: dfi
Out[2]:
A B
0 0 1
1 2 3
2 4 5
In [3]: dfi.loc[:,'C'] = dfi.loc[:,'A']
In [4]: dfi
Out[4]:
A B C
0 0 1 0
1 2 3 2
2 4 5 4
In [5]: dfi.loc[3] = 5
In [6]: dfi
Out[6]:
A B C
0 0 1 0
1 2 3 2
2 4 5 4
3 5 5 5
For the sake of a Pythonic way:
res = pd.DataFrame(columns=('lib', 'qty1', 'qty2'))
res = res.append([{'qty1':10.0}], ignore_index=True)
print(res.head())
lib qty1 qty2
0 NaN 10.0 NaN
You can also build up a list of lists and convert it to a dataframe -
import pandas as pd
columns = ['i','double','square']
rows = []
for i in range(6):
row = [i, i*2, i*i]
rows.append(row)
df = pd.DataFrame(rows, columns=columns)
giving
i double square
0 0 0 0
1 1 2 1
2 2 4 4
3 3 6 9
4 4 8 16
5 5 10 25
If you always want to add a new row at the end, use this:
df.loc[len(df)] = ['name5', 9, 0]
I figured out a simple and nice way:
>>> df
A B C
one 1 2 3
>>> df.loc["two"] = [4,5,6]
>>> df
A B C
one 1 2 3
two 4 5 6
Note the caveat with performance as noted in the comments.
This is not an answer to the OP question, but a toy example to illustrate ShikharDua's answer which I found very useful.
While this fragment is trivial, in the actual data I had 1,000s of rows, and many columns, and I wished to be able to group by different columns and then perform the statistics below for more than one target column. So having a reliable method for building the data frame one row at a time was a great convenience. Thank you ShikharDua!
import pandas as pd
BaseData = pd.DataFrame({ 'Customer' : ['Acme','Mega','Acme','Acme','Mega','Acme'],
'Territory' : ['West','East','South','West','East','South'],
'Product' : ['Econ','Luxe','Econ','Std','Std','Econ']})
BaseData
columns = ['Customer','Num Unique Products', 'List Unique Products']
rows_list=[]
for name, group in BaseData.groupby('Customer'):
RecordtoAdd={} #initialise an empty dict
RecordtoAdd.update({'Customer' : name}) #
RecordtoAdd.update({'Num Unique Products' : len(pd.unique(group['Product']))})
RecordtoAdd.update({'List Unique Products' : pd.unique(group['Product'])})
rows_list.append(RecordtoAdd)
AnalysedData = pd.DataFrame(rows_list)
print('Base Data : \n',BaseData,'\n\n Analysed Data : \n',AnalysedData)
You can use a generator object to create a Dataframe, which will be more memory efficient over the list.
num = 10
# Generator function to generate generator object
def numgen_func(num):
for i in range(num):
yield ('name_{}'.format(i), (i*i), (i*i*i))
# Generator expression to generate generator object (Only once data get populated, can not be re used)
numgen_expression = (('name_{}'.format(i), (i*i), (i*i*i)) for i in range(num) )
df = pd.DataFrame(data=numgen_func(num), columns=('lib', 'qty1', 'qty2'))
To add raw to existing DataFrame you can use append method.
df = df.append([{ 'lib': "name_20", 'qty1': 20, 'qty2': 400 }])
Instead of a list of dictionaries as in ShikharDua's answer (row-based), we can also represent our table as a dictionary of lists (column-based), where each list stores one column in row-order, given we know our columns beforehand. At the end we construct our DataFrame once.
In both cases, the dictionary keys are always the column names. Row order is stored implicitly as order in a list. For c columns and n rows, this uses one dictionary of c lists, versus one list of n dictionaries. The list-of-dictionaries method has each dictionary storing all keys redundantly and requires creating a new dictionary for every row. Here we only append to lists, which overall is the same time complexity (adding entries to list and dictionary are both amortized constant time) but may have less overhead due to being a simple operation.
# Current data
data = {"Animal":["cow", "horse"], "Color":["blue", "red"]}
# Adding a new row (be careful to ensure every column gets another value)
data["Animal"].append("mouse")
data["Color"].append("black")
# At the end, construct our DataFrame
df = pd.DataFrame(data)
# Animal Color
# 0 cow blue
# 1 horse red
# 2 mouse black
Create a new record (data frame) and add to old_data_frame.
Pass a list of values and the corresponding column names to create a new_record (data_frame):
new_record = pd.DataFrame([[0, 'abcd', 0, 1, 123]], columns=['a', 'b', 'c', 'd', 'e'])
old_data_frame = pd.concat([old_data_frame, new_record])
Here is the way to add/append a row in a Pandas DataFrame:
def add_row(df, row):
df.loc[-1] = row
df.index = df.index + 1
return df.sort_index()
add_row(df, [1,2,3])
It can be used to insert/append a row in an empty or populated Pandas DataFrame.
If you want to add a row at the end, append it as a list:
valuestoappend = [va1, val2, val3]
res = res.append(pd.Series(valuestoappend, index = ['lib', 'qty1', 'qty2']), ignore_index = True)
Another way to do it (probably not very performant):
# add a row
def add_row(df, row):
colnames = list(df.columns)
ncol = len(colnames)
assert ncol == len(row), "Length of row must be the same as width of DataFrame: %s" % row
return df.append(pd.DataFrame([row], columns=colnames))
You can also enhance the DataFrame class like this:
import pandas as pd
def add_row(self, row):
self.loc[len(self.index)] = row
pd.DataFrame.add_row = add_row
All you need is loc[df.shape[0]] or loc[len(df)]
# Assuming your df has 4 columns (str, int, str, bool)
df.loc[df.shape[0]] = ['col1Value', 100, 'col3Value', False]
or
df.loc[len(df)] = ['col1Value', 100, 'col3Value', False]
You can concatenate two DataFrames for this. I basically came across this problem to add a new row to an existing DataFrame with a character index (not numeric).
So, I input the data for a new row in a duct() and index in a list.
new_dict = {put input for new row here}
new_list = [put your index here]
new_df = pd.DataFrame(data=new_dict, index=new_list)
df = pd.concat([existing_df, new_df])
initial_data = {'lib': np.array([1,2,3,4]), 'qty1': [1,2,3,4], 'qty2': [1,2,3,4]}
df = pd.DataFrame(initial_data)
df
lib qty1 qty2
0 1 1 1
1 2 2 2
2 3 3 3
3 4 4 4
val_1 = [10]
val_2 = [14]
val_3 = [20]
df.append(pd.DataFrame({'lib': val_1, 'qty1': val_2, 'qty2': val_3}))
lib qty1 qty2
0 1 1 1
1 2 2 2
2 3 3 3
3 4 4 4
0 10 14 20
You can use a for loop to iterate through values or can add arrays of values.
val_1 = [10, 11, 12, 13]
val_2 = [14, 15, 16, 17]
val_3 = [20, 21, 22, 43]
df.append(pd.DataFrame({'lib': val_1, 'qty1': val_2, 'qty2': val_3}))
lib qty1 qty2
0 1 1 1
1 2 2 2
2 3 3 3
3 4 4 4
0 10 14 20
1 11 15 21
2 12 16 22
3 13 17 43
Make it simple. By taking a list as input which will be appended as a row in the data-frame:
import pandas as pd
res = pd.DataFrame(columns=('lib', 'qty1', 'qty2'))
for i in range(5):
res_list = list(map(int, input().split()))
res = res.append(pd.Series(res_list, index=['lib', 'qty1', 'qty2']), ignore_index=True)
pandas.DataFrame.append
DataFrame.append(self, other, ignore_index=False, verify_integrity=False, sort=False) → 'DataFrame'
Code
df = pd.DataFrame([[1, 2], [3, 4]], columns=list('AB'))
df2 = pd.DataFrame([[5, 6], [7, 8]], columns=list('AB'))
df.append(df2)
With ignore_index set to True:
df.append(df2, ignore_index=True)
If you have a data frame df and want to add a list new_list as a new row to df, you can simply do:
df.loc[len(df)] = new_list
If you want to add a new data frame new_df under data frame df, then you can use:
df.append(new_df)
We often see the construct df.loc[subscript] = … to assign to one DataFrame row. Mikhail_Sam posted benchmarks containing, among others, this construct as well as the method using dict and create DataFrame in the end. He found the latter to be the fastest by far.
But if we replace the df3.loc[i] = … (with preallocated DataFrame) in his code with df3.values[i] = …, the outcome changes significantly, in that that method performs similar to the one using dict. So we should more often take the use of df.values[subscript] = … into consideration. However note that .values takes a zero-based subscript, which may be different from the DataFrame.index.
Before going to add a row, we have to convert the dataframe to a dictionary. There you can see the keys as columns in the dataframe and the values of the columns are again stored in the dictionary, but there the key for every column is the index number in the dataframe.
That idea makes me to write the below code.
df2 = df.to_dict()
values = ["s_101", "hyderabad", 10, 20, 16, 13, 15, 12, 12, 13, 25, 26, 25, 27, "good", "bad"] # This is the total row that we are going to add
i = 0
for x in df.columns: # Here df.columns gives us the main dictionary key
df2[x][101] = values[i] # Here the 101 is our index number. It is also the key of the sub dictionary
i += 1
If all data in your Dataframe has the same dtype you might use a NumPy array. You can write rows directly into the predefined array and convert it to a dataframe at the end.
It seems to be even faster than converting a list of dicts.
import pandas as pd
import numpy as np
from string import ascii_uppercase
startTime = time.perf_counter()
numcols, numrows = 5, 10000
npdf = np.ones((numrows, numcols))
for row in range(numrows):
npdf[row, 0:] = np.random.randint(0, 100, (1, numcols))
df5 = pd.DataFrame(npdf, columns=list(ascii_uppercase[:numcols]))
print('Elapsed time: {:6.3f} seconds for {:d} rows'.format(time.perf_counter() - startTime, numOfRows))
print(df5.shape)
This code snippet uses a list of dictionaries to update the data frame. It adds on to ShikharDua's and Mikhail_Sam's answers.
import pandas as pd
colour = ["red", "big", "tasty"]
fruits = ["apple", "banana", "cherry"]
dict1={}
feat_list=[]
for x in colour:
for y in fruits:
# print(x, y)
dict1 = dict([('x',x),('y',y)])
# print(f'dict 1 {dict1}')
feat_list.append(dict1)
# print(f'feat_list {feat_list}')
feat_df=pd.DataFrame(feat_list)
feat_df.to_csv('feat1.csv')

Merge data frames based on column with different rows

I have multiple csv files that I read into individual data frames based on their name in the directory, like so
# ask user for path
path = input('Enter the path for the csv files: ')
os.chdir(path)
# loop over filenames and read into individual dataframes
for fname in os.listdir(path):
if fname.endswith('Demo.csv'):
demoRaw = pd.read_csv(fname, encoding = 'utf-8')
if fname.endswith('Key2.csv'):
keyRaw = pd.read_csv(fname, encoding = 'utf-8')
Then I filter to only keep certain columns
# filter to keep desired columns only
demo = demoRaw.filter(['Key', 'Sex', 'Race', 'Age'], axis=1)
key = keyRaw.filter(['Key', 'Key', 'Age'], axis=1)
Then I create a list of the above dataframes and use reduce to merge them on Key
# create list of data frames for combined sheet
dfs = [demo, key]
# merge the list of data frames on the Key
combined = reduce(lambda left,right: pd.merge(left,right,on='Key'), dfs)
Then I drop the auto generated column, create an Excel writer and write to a csv
# drop the auto generated index colulmn
combined.set_index('RecordKey', inplace=True)
# create a Pandas Excel writer using XlsxWriter as the engine.
writer = pd.ExcelWriter('final.xlsx', engine='xlsxwriter')
# write to csv
combined.to_excel(writer, sheet_name='Combined')
meds.to_excel(writer, sheet_name='Meds')
# Close the Pandas Excel writer and output the Excel file.
writer.save()
The problem is some files have keys that aren't in others. For example
Demo file
Key Sex Race Age
1 M W 52
2 F B 25
3 M L 78
Key file
Key Key2 Age
1 7325 52
2 4783 25
3 1367 78
4 9435 21
5 7247 65
Right now, it will only include rows if there is a matching key in each (in other words it just leaves out the rows with keys not in the other files). How can I combine all rows from all files, even if keys don't match? So the end result will look like this
Key Sex Race Age Key2 Age
1 M W 52 7325 52
2 F B 25 4783 25
3 M L 78 1367 78
4 9435 21
5 7247 65
I don't care if the empty cells are blanks, NaN, #N/A, etc. Just as long as I can identify them.
Replace combined = reduce(lambda left,right: pd.merge(left,right,on='Key'), dfs) With: combined=pd.merge(demo,key, how='outer', on='Key') You will have to specificy the 'outer' to join both the full table of Key and Demo

Appending values to a column in a loop

I have various files containing data. I want to extract one specific column from each file and create a new dataframe with one column containing all the extracted data.
So for example I have 3 files:
A B C
1 2 3
4 5 6
A B C
7 8 9
8 7 6
A B C
5 4 3
2 1 0
The new dataframe should only contain the values from column C:
C
3
6
9
6
3
0
So the column of the first file should be copied to the new dataframe, the column from the second file should be appendend to the new dataframe.
My code looks like this so far:
import pandas as pd
import glob
for filename in glob.glob('*.dat'):
df= pd.read_csv(filename, delimiter="\t", header=6)
df1= df["Bias"]
print(df)
Now df1 is overwritten in each loop step. Would it be a good idea to create a temporary dataframe in each loop step and then copy the data to the new dataframe?
Any input is appreciated!
Use list comprehension or for loop with append for list of DataFrames and if need only some columns add parameter usecols, last concat all together for big DataFrame:
dfs = [pd.read_csv(f, delimiter="\t", header=6, usecols=['C']) for f in glob.glob('*.dat')]
Or:
dfs = []
for filename in glob.glob('*.dat'):
df = pd.read_csv(filename, delimiter="\t", header=6, usecols=['C'])
#if need all columns
#df = pd.read_csv(filename, delimiter="\t", header=6)
dfs.append(df)
df = pd.concat(dfs, ignore_index=True)

Accumalate column through pandas

I have multiple tab delimited files, all having same entries. I intend to read each file choose first column as index. My final table will have first column as index mapped against last column from all the files. For this, I wrote a pandas code but not a great ones. Is there an alternate way to do this ?
import pandas as pd
df1 = pd.read_csv("FB_test.tsv",sep='\t')
df1_idx = df1.set_index('target_id')
df1_idx.drop(df1_idx[['length','eff_length','est_counts']],inplace=True, axis=1)
print(df1_idx)
df2 = pd.read_csv("Myc_test.tsv",sep='\t')
df2_idx = df2.set_index('target_id')
df2_idx.drop(df2_idx[['length','eff_length','est_counts']],inplace=True, axis=1)
print(df2_idx)
frames = [df1_idx, df2_idx]
results = pd.concat(frames, axis=1)
results
The output it generated was,
tpm
target_id
A 0
B 0
C 0
D 0
E 0
tpm
target_id
A 1
B 1
C 1
D 1
E 1
Out[18]:
target_id tpm tpm
A 0 1
B 0 1
C 0 1
D 0 1
E 0 1
How to loop it so that, I read each file and achieve this same output ?
Thanks,
AP
I think you can use parameters index_col and usecols in read_csv with list comprehension. But get duplicates columns names (so is problem for selecting), so better is add parameter keys to concat - after converting Multiindex get nice unique column names:
files = ["FB_test.tsv", "Myc_test.tsv"]
dfs = [pd.read_csv(f,sep='\t', index_col=['target_id'], usecols=['target_id','tpm'])
for f in files]
results = pd.concat(dfs, axis=1, keys=('a','b'))
results.columns = results.columns.map('_'.join)
results = results.reset_index()
print (results)
target_id a_tpm b_tpm
0 A 0 1
1 B 0 1
2 C 0 1
3 D 0 1
4 E 0 1
To clean the code and use a looping mechanism, you can put both your file names and the columns you are dropping in two separate lists, and then use list comprehension on the file names to import each dataset. Subsequently, you concatenate the output of the list comprehension into one dataframe:
import pandas as pd
drop_cols = ['length','eff_length','est_counts']
filenames = ["FB_test.tsv", "Myc_test.tsv"]
results = pd.concat([pd.read_csv(filename,sep='\t').set_index('target_id').drop(drop_cols, axis=1) for filename in filenames], axis=1)
I hope this helps.

Pandas writing in csv file as columns not rows-Python

This is my code:
import os
file=[]
directory ='/Users/xxxx/Documents/sample/'
for i in os.listdir(directory):
file.append(i)
Com = list(file)
df=pd.DataFrame(data=Com)
df.to_csv('com.csv', index=False, header=True)
print('done')
at the moment I am getting all the values for i in one column as a row header. Does anyone know how to make each i value in one row as a column header?
You need to transpose the df first using .T prior to writing out to csv:
In [44]:
l = list('abc')
df = pd.DataFrame(l)
df
Out[44]:
0
0 a
1 b
2 c
compare with:
In [45]:
df = pd.DataFrame(l).T
df
Out[45]:
0 1 2
0 a b c

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