How to replace number in csv with a string with python - python

I am trying to fix the first row of a CSV file. If column name in header starts from anything other than a-z, NUM has to be prepended. The following code fixes the special characters in each column of the first row but somehow can't get the !a-z.
path = ('test.csv')
for fname in glob.glob(path):
with open(fname, newline='') as f:
reader = csv.reader(f)
header = next(reader)
header = [column.replace ('-','_') for column in header]
header = [column.replace ('[!a-z]','NUM') for column in header]
what am I doing wrong. Please provide suggestions.
Thanks

You can do it like this.
# csv file:
# 2Hello, ?WORLD
# 1, 2
import csv
with open("test.csv", newline='') as f:
reader = csv.reader(f)
header = next(reader)
print("Original header", header)
header = [("NUM" + header[indx][1::]) for indx in range(len(header)) if not header[indx][0].isalpha()]
print("Modified header", header)
Output:
Original header ['2HELLO', '?WORLD']
Modified header ['NUMHELLO', 'NUMWORLD']
The above list comprehension is equivalent to the following for loop:
for indx in range(len(header)):
if not header[indx][0].isalpha():
header[indx] = "NUM" + header[indx][1::]
If you want to replace only numbers, then use the following:
if header[indx][0].isdigit():
You can modify this according to your requirements in case if it changes based on many relevant string functions.
https://docs.python.org/2/library/string.html

I believe you would want to replace the 'column.replace' portion with something along these lines:
re.sub(r'[!a-z]', 'NUM', column)
The full documentation reference is here for specifics: https://docs.python.org/2/library/re.html
https://www.regular-expressions.info/python.html

Since you said you want to prepend 'NUM', you could do something like this (which could be more efficient, but this shows the basic idea).
import string
column = '123'
if column[0] not in string.ascii_lowercase:
column = 'NUM' + column
# column is now 'NUM123'

Related

How to take a specific string from a csv file into a list - python 3.4

I have a csv list with 3 rows but i only want to take the first string from each of these rows and append them to a new list.
names = list()
a = open(sf, 'r')
for row in a:
names.append(row[0])
print (names)
The output I get from this is only the first letter of each of the strings. How do I go about it so I get the entire string as they all vary in lengths. Thanks in advance
try looking for the comma in the csv:
names = list()
a = open(sf, 'r')
for row in a:
place = row.find(',')
names.append(row[:place])
print(names)
Use csv module to do it.
import csv
names = []
a = open(sf, 'r')
reader = csv.reader(a, delimiter=',')
for row in reader:
names.append(row[0])
print names
Another way is like this:
names = []
with open(sf, 'r') as fi:
for row in fi:
names.append(row.split(',')[0])
print(names)
Here you split the row on delimiter of your csv (here on comma) and append the first element to the names.
If you would rather split on whitespace, you can use split() without argument.

Trying to convert a CSV file to int in Python [duplicate]

I am asking Python to print the minimum number from a column of CSV data, but the top row is the column number, and I don't want Python to take the top row into account. How can I make sure Python ignores the first line?
This is the code so far:
import csv
with open('all16.csv', 'rb') as inf:
incsv = csv.reader(inf)
column = 1
datatype = float
data = (datatype(column) for row in incsv)
least_value = min(data)
print least_value
Could you also explain what you are doing, not just give the code? I am very very new to Python and would like to make sure I understand everything.
You could use an instance of the csv module's Sniffer class to deduce the format of a CSV file and detect whether a header row is present along with the built-in next() function to skip over the first row only when necessary:
import csv
with open('all16.csv', 'r', newline='') as file:
has_header = csv.Sniffer().has_header(file.read(1024))
file.seek(0) # Rewind.
reader = csv.reader(file)
if has_header:
next(reader) # Skip header row.
column = 1
datatype = float
data = (datatype(row[column]) for row in reader)
least_value = min(data)
print(least_value)
Since datatype and column are hardcoded in your example, it would be slightly faster to process the row like this:
data = (float(row[1]) for row in reader)
Note: the code above is for Python 3.x. For Python 2.x use the following line to open the file instead of what is shown:
with open('all16.csv', 'rb') as file:
To skip the first line just call:
next(inf)
Files in Python are iterators over lines.
Borrowed from python cookbook,
A more concise template code might look like this:
import csv
with open('stocks.csv') as f:
f_csv = csv.reader(f)
headers = next(f_csv)
for row in f_csv:
# Process row ...
In a similar use case I had to skip annoying lines before the line with my actual column names. This solution worked nicely. Read the file first, then pass the list to csv.DictReader.
with open('all16.csv') as tmp:
# Skip first line (if any)
next(tmp, None)
# {line_num: row}
data = dict(enumerate(csv.DictReader(tmp)))
You would normally use next(incsv) which advances the iterator one row, so you skip the header. The other (say you wanted to skip 30 rows) would be:
from itertools import islice
for row in islice(incsv, 30, None):
# process
use csv.DictReader instead of csv.Reader.
If the fieldnames parameter is omitted, the values in the first row of the csvfile will be used as field names. you would then be able to access field values using row["1"] etc
Python 2.x
csvreader.next()
Return the next row of the reader’s iterable object as a list, parsed
according to the current dialect.
csv_data = csv.reader(open('sample.csv'))
csv_data.next() # skip first row
for row in csv_data:
print(row) # should print second row
Python 3.x
csvreader.__next__()
Return the next row of the reader’s iterable object as a list (if the
object was returned from reader()) or a dict (if it is a DictReader
instance), parsed according to the current dialect. Usually you should
call this as next(reader).
csv_data = csv.reader(open('sample.csv'))
csv_data.__next__() # skip first row
for row in csv_data:
print(row) # should print second row
The documentation for the Python 3 CSV module provides this example:
with open('example.csv', newline='') as csvfile:
dialect = csv.Sniffer().sniff(csvfile.read(1024))
csvfile.seek(0)
reader = csv.reader(csvfile, dialect)
# ... process CSV file contents here ...
The Sniffer will try to auto-detect many things about the CSV file. You need to explicitly call its has_header() method to determine whether the file has a header line. If it does, then skip the first row when iterating the CSV rows. You can do it like this:
if sniffer.has_header():
for header_row in reader:
break
for data_row in reader:
# do something with the row
this might be a very old question but with pandas we have a very easy solution
import pandas as pd
data=pd.read_csv('all16.csv',skiprows=1)
data['column'].min()
with skiprows=1 we can skip the first row then we can find the least value using data['column'].min()
The new 'pandas' package might be more relevant than 'csv'. The code below will read a CSV file, by default interpreting the first line as the column header and find the minimum across columns.
import pandas as pd
data = pd.read_csv('all16.csv')
data.min()
Because this is related to something I was doing, I'll share here.
What if we're not sure if there's a header and you also don't feel like importing sniffer and other things?
If your task is basic, such as printing or appending to a list or array, you could just use an if statement:
# Let's say there's 4 columns
with open('file.csv') as csvfile:
csvreader = csv.reader(csvfile)
# read first line
first_line = next(csvreader)
# My headers were just text. You can use any suitable conditional here
if len(first_line) == 4:
array.append(first_line)
# Now we'll just iterate over everything else as usual:
for row in csvreader:
array.append(row)
Well, my mini wrapper library would do the job as well.
>>> import pyexcel as pe
>>> data = pe.load('all16.csv', name_columns_by_row=0)
>>> min(data.column[1])
Meanwhile, if you know what header column index one is, for example "Column 1", you can do this instead:
>>> min(data.column["Column 1"])
For me the easiest way to go is to use range.
import csv
with open('files/filename.csv') as I:
reader = csv.reader(I)
fulllist = list(reader)
# Starting with data skipping header
for item in range(1, len(fulllist)):
# Print each row using "item" as the index value
print (fulllist[item])
I would convert csvreader to list, then pop the first element
import csv
with open(fileName, 'r') as csvfile:
csvreader = csv.reader(csvfile)
data = list(csvreader) # Convert to list
data.pop(0) # Removes the first row
for row in data:
print(row)
I would use tail to get rid of the unwanted first line:
tail -n +2 $INFIL | whatever_script.py
just add [1:]
example below:
data = pd.read_csv("/Users/xyz/Desktop/xyxData/xyz.csv", sep=',', header=None)**[1:]**
that works for me in iPython
Python 3.X
Handles UTF8 BOM + HEADER
It was quite frustrating that the csv module could not easily get the header, there is also a bug with the UTF-8 BOM (first char in file).
This works for me using only the csv module:
import csv
def read_csv(self, csv_path, delimiter):
with open(csv_path, newline='', encoding='utf-8') as f:
# https://bugs.python.org/issue7185
# Remove UTF8 BOM.
txt = f.read()[1:]
# Remove header line.
header = txt.splitlines()[:1]
lines = txt.splitlines()[1:]
# Convert to list.
csv_rows = list(csv.reader(lines, delimiter=delimiter))
for row in csv_rows:
value = row[INDEX_HERE]
Simple Solution is to use csv.DictReader()
import csv
def read_csv(file): with open(file, 'r') as file:
reader = csv.DictReader(file)
for row in reader:
print(row["column_name"]) # Replace the name of column header.

Reading column names alone in a csv file

I have a csv file with the following columns:
id,name,age,sex
Followed by a lot of values for the above columns.
I am trying to read the column names alone and put them inside a list.
I am using Dictreader and this gives out the correct details:
with open('details.csv') as csvfile:
i=["name","age","sex"]
re=csv.DictReader(csvfile)
for row in re:
for x in i:
print row[x]
But what I want to do is, I need the list of columns, ("i" in the above case)to be automatically parsed with the input csv than hardcoding them inside a list.
with open('details.csv') as csvfile:
rows=iter(csv.reader(csvfile)).next()
header=rows[1:]
re=csv.DictReader(csvfile)
for row in re:
print row
for x in header:
print row[x]
This gives out an error
Keyerrror:'name'
in the line print row[x]. Where am I going wrong? Is it possible to fetch the column names using Dictreader?
Though you already have an accepted answer, I figured I'd add this for anyone else interested in a different solution-
Python's DictReader object in the CSV module (as of Python 2.6 and above) has a public attribute called fieldnames.
https://docs.python.org/3.4/library/csv.html#csv.csvreader.fieldnames
An implementation could be as follows:
import csv
with open('C:/mypath/to/csvfile.csv', 'r') as f:
d_reader = csv.DictReader(f)
#get fieldnames from DictReader object and store in list
headers = d_reader.fieldnames
for line in d_reader:
#print value in MyCol1 for each row
print(line['MyCol1'])
In the above, d_reader.fieldnames returns a list of your headers (assuming the headers are in the top row).
Which allows...
>>> print(headers)
['MyCol1', 'MyCol2', 'MyCol3']
If your headers are in, say the 2nd row (with the very top row being row 1), you could do as follows:
import csv
with open('C:/mypath/to/csvfile.csv', 'r') as f:
#you can eat the first line before creating DictReader.
#if no "fieldnames" param is passed into
#DictReader object upon creation, DictReader
#will read the upper-most line as the headers
f.readline()
d_reader = csv.DictReader(f)
headers = d_reader.fieldnames
for line in d_reader:
#print value in MyCol1 for each row
print(line['MyCol1'])
You can read the header by using the next() function which return the next row of the reader’s iterable object as a list. then you can add the content of the file to a list.
import csv
with open("C:/path/to/.filecsv", "rb") as f:
reader = csv.reader(f)
i = reader.next()
rest = list(reader)
Now i has the column's names as a list.
print i
>>>['id', 'name', 'age', 'sex']
Also note that reader.next() does not work in python 3. Instead use the the inbuilt next() to get the first line of the csv immediately after reading like so:
import csv
with open("C:/path/to/.filecsv", "rb") as f:
reader = csv.reader(f)
i = next(reader)
print(i)
>>>['id', 'name', 'age', 'sex']
The csv.DictReader object exposes an attribute called fieldnames, and that is what you'd use. Here's example code, followed by input and corresponding output:
import csv
file = "/path/to/file.csv"
with open(file, mode='r', encoding='utf-8') as f:
reader = csv.DictReader(f, delimiter=',')
for row in reader:
print([col + '=' + row[col] for col in reader.fieldnames])
Input file contents:
col0,col1,col2,col3,col4,col5,col6,col7,col8,col9
00,01,02,03,04,05,06,07,08,09
10,11,12,13,14,15,16,17,18,19
20,21,22,23,24,25,26,27,28,29
30,31,32,33,34,35,36,37,38,39
40,41,42,43,44,45,46,47,48,49
50,51,52,53,54,55,56,57,58,59
60,61,62,63,64,65,66,67,68,69
70,71,72,73,74,75,76,77,78,79
80,81,82,83,84,85,86,87,88,89
90,91,92,93,94,95,96,97,98,99
Output of print statements:
['col0=00', 'col1=01', 'col2=02', 'col3=03', 'col4=04', 'col5=05', 'col6=06', 'col7=07', 'col8=08', 'col9=09']
['col0=10', 'col1=11', 'col2=12', 'col3=13', 'col4=14', 'col5=15', 'col6=16', 'col7=17', 'col8=18', 'col9=19']
['col0=20', 'col1=21', 'col2=22', 'col3=23', 'col4=24', 'col5=25', 'col6=26', 'col7=27', 'col8=28', 'col9=29']
['col0=30', 'col1=31', 'col2=32', 'col3=33', 'col4=34', 'col5=35', 'col6=36', 'col7=37', 'col8=38', 'col9=39']
['col0=40', 'col1=41', 'col2=42', 'col3=43', 'col4=44', 'col5=45', 'col6=46', 'col7=47', 'col8=48', 'col9=49']
['col0=50', 'col1=51', 'col2=52', 'col3=53', 'col4=54', 'col5=55', 'col6=56', 'col7=57', 'col8=58', 'col9=59']
['col0=60', 'col1=61', 'col2=62', 'col3=63', 'col4=64', 'col5=65', 'col6=66', 'col7=67', 'col8=68', 'col9=69']
['col0=70', 'col1=71', 'col2=72', 'col3=73', 'col4=74', 'col5=75', 'col6=76', 'col7=77', 'col8=78', 'col9=79']
['col0=80', 'col1=81', 'col2=82', 'col3=83', 'col4=84', 'col5=85', 'col6=86', 'col7=87', 'col8=88', 'col9=89']
['col0=90', 'col1=91', 'col2=92', 'col3=93', 'col4=94', 'col5=95', 'col6=96', 'col7=97', 'col8=98', 'col9=99']
How about
with open(csv_input_path + file, 'r') as ft:
header = ft.readline() # read only first line; returns string
header_list = header.split(',') # returns list
I am assuming your input file is CSV format.
If using pandas, it takes more time if the file is big size because it loads the entire data as the dataset.
I am just mentioning how to get all the column names from a csv file.
I am using pandas library.
First we read the file.
import pandas as pd
file = pd.read_csv('details.csv')
Then, in order to just get all the column names as a list from input file use:-
columns = list(file.head(0))
Thanking Daniel Jimenez for his perfect solution to fetch column names alone from my csv, I extend his solution to use DictReader so we can iterate over the rows using column names as indexes. Thanks Jimenez.
with open('myfile.csv') as csvfile:
rest = []
with open("myfile.csv", "rb") as f:
reader = csv.reader(f)
i = reader.next()
i=i[1:]
re=csv.DictReader(csvfile)
for row in re:
for x in i:
print row[x]
here is the code to print only the headers or columns of the csv file.
import csv
HEADERS = next(csv.reader(open('filepath.csv')))
print (HEADERS)
Another method with pandas
import pandas as pd
HEADERS = list(pd.read_csv('filepath.csv').head(0))
print (HEADERS)
import pandas as pd
data = pd.read_csv("data.csv")
cols = data.columns
I literally just wanted the first row of my data which are the headers I need and didn't want to iterate over all my data to get them, so I just did this:
with open(data, 'r', newline='') as csvfile:
t = 0
for i in csv.reader(csvfile, delimiter=',', quotechar='|'):
if t > 0:
break
else:
dbh = i
t += 1
Using pandas is also an option.
But instead of loading the full file in memory, you can retrieve only the first chunk of it to get the field names by using iterator.
import pandas as pd
file = pd.read_csv('details.csv'), iterator=True)
column_names_full=file.get_chunk(1)
column_names=[column for column in column_names_full]
print column_names

How to ignore the first line of data when processing CSV data?

I am asking Python to print the minimum number from a column of CSV data, but the top row is the column number, and I don't want Python to take the top row into account. How can I make sure Python ignores the first line?
This is the code so far:
import csv
with open('all16.csv', 'rb') as inf:
incsv = csv.reader(inf)
column = 1
datatype = float
data = (datatype(column) for row in incsv)
least_value = min(data)
print least_value
Could you also explain what you are doing, not just give the code? I am very very new to Python and would like to make sure I understand everything.
You could use an instance of the csv module's Sniffer class to deduce the format of a CSV file and detect whether a header row is present along with the built-in next() function to skip over the first row only when necessary:
import csv
with open('all16.csv', 'r', newline='') as file:
has_header = csv.Sniffer().has_header(file.read(1024))
file.seek(0) # Rewind.
reader = csv.reader(file)
if has_header:
next(reader) # Skip header row.
column = 1
datatype = float
data = (datatype(row[column]) for row in reader)
least_value = min(data)
print(least_value)
Since datatype and column are hardcoded in your example, it would be slightly faster to process the row like this:
data = (float(row[1]) for row in reader)
Note: the code above is for Python 3.x. For Python 2.x use the following line to open the file instead of what is shown:
with open('all16.csv', 'rb') as file:
To skip the first line just call:
next(inf)
Files in Python are iterators over lines.
Borrowed from python cookbook,
A more concise template code might look like this:
import csv
with open('stocks.csv') as f:
f_csv = csv.reader(f)
headers = next(f_csv)
for row in f_csv:
# Process row ...
In a similar use case I had to skip annoying lines before the line with my actual column names. This solution worked nicely. Read the file first, then pass the list to csv.DictReader.
with open('all16.csv') as tmp:
# Skip first line (if any)
next(tmp, None)
# {line_num: row}
data = dict(enumerate(csv.DictReader(tmp)))
You would normally use next(incsv) which advances the iterator one row, so you skip the header. The other (say you wanted to skip 30 rows) would be:
from itertools import islice
for row in islice(incsv, 30, None):
# process
use csv.DictReader instead of csv.Reader.
If the fieldnames parameter is omitted, the values in the first row of the csvfile will be used as field names. you would then be able to access field values using row["1"] etc
Python 2.x
csvreader.next()
Return the next row of the reader’s iterable object as a list, parsed
according to the current dialect.
csv_data = csv.reader(open('sample.csv'))
csv_data.next() # skip first row
for row in csv_data:
print(row) # should print second row
Python 3.x
csvreader.__next__()
Return the next row of the reader’s iterable object as a list (if the
object was returned from reader()) or a dict (if it is a DictReader
instance), parsed according to the current dialect. Usually you should
call this as next(reader).
csv_data = csv.reader(open('sample.csv'))
csv_data.__next__() # skip first row
for row in csv_data:
print(row) # should print second row
The documentation for the Python 3 CSV module provides this example:
with open('example.csv', newline='') as csvfile:
dialect = csv.Sniffer().sniff(csvfile.read(1024))
csvfile.seek(0)
reader = csv.reader(csvfile, dialect)
# ... process CSV file contents here ...
The Sniffer will try to auto-detect many things about the CSV file. You need to explicitly call its has_header() method to determine whether the file has a header line. If it does, then skip the first row when iterating the CSV rows. You can do it like this:
if sniffer.has_header():
for header_row in reader:
break
for data_row in reader:
# do something with the row
this might be a very old question but with pandas we have a very easy solution
import pandas as pd
data=pd.read_csv('all16.csv',skiprows=1)
data['column'].min()
with skiprows=1 we can skip the first row then we can find the least value using data['column'].min()
The new 'pandas' package might be more relevant than 'csv'. The code below will read a CSV file, by default interpreting the first line as the column header and find the minimum across columns.
import pandas as pd
data = pd.read_csv('all16.csv')
data.min()
Because this is related to something I was doing, I'll share here.
What if we're not sure if there's a header and you also don't feel like importing sniffer and other things?
If your task is basic, such as printing or appending to a list or array, you could just use an if statement:
# Let's say there's 4 columns
with open('file.csv') as csvfile:
csvreader = csv.reader(csvfile)
# read first line
first_line = next(csvreader)
# My headers were just text. You can use any suitable conditional here
if len(first_line) == 4:
array.append(first_line)
# Now we'll just iterate over everything else as usual:
for row in csvreader:
array.append(row)
Well, my mini wrapper library would do the job as well.
>>> import pyexcel as pe
>>> data = pe.load('all16.csv', name_columns_by_row=0)
>>> min(data.column[1])
Meanwhile, if you know what header column index one is, for example "Column 1", you can do this instead:
>>> min(data.column["Column 1"])
For me the easiest way to go is to use range.
import csv
with open('files/filename.csv') as I:
reader = csv.reader(I)
fulllist = list(reader)
# Starting with data skipping header
for item in range(1, len(fulllist)):
# Print each row using "item" as the index value
print (fulllist[item])
I would convert csvreader to list, then pop the first element
import csv
with open(fileName, 'r') as csvfile:
csvreader = csv.reader(csvfile)
data = list(csvreader) # Convert to list
data.pop(0) # Removes the first row
for row in data:
print(row)
I would use tail to get rid of the unwanted first line:
tail -n +2 $INFIL | whatever_script.py
just add [1:]
example below:
data = pd.read_csv("/Users/xyz/Desktop/xyxData/xyz.csv", sep=',', header=None)**[1:]**
that works for me in iPython
Python 3.X
Handles UTF8 BOM + HEADER
It was quite frustrating that the csv module could not easily get the header, there is also a bug with the UTF-8 BOM (first char in file).
This works for me using only the csv module:
import csv
def read_csv(self, csv_path, delimiter):
with open(csv_path, newline='', encoding='utf-8') as f:
# https://bugs.python.org/issue7185
# Remove UTF8 BOM.
txt = f.read()[1:]
# Remove header line.
header = txt.splitlines()[:1]
lines = txt.splitlines()[1:]
# Convert to list.
csv_rows = list(csv.reader(lines, delimiter=delimiter))
for row in csv_rows:
value = row[INDEX_HERE]
Simple Solution is to use csv.DictReader()
import csv
def read_csv(file): with open(file, 'r') as file:
reader = csv.DictReader(file)
for row in reader:
print(row["column_name"]) # Replace the name of column header.

How to avoid splitting on a delimiter if it appears inside quotes?

I have problem in splitting data. I have data as follows in CSV file:
"a";"b";"c;d";"e"
The problem is when I used line.split(";") function, it splits even between c and d. I don't want c and d to be separated. Later I need to store these four values in four different columns in a table, but using this function I get five different columns.
I want the results to be "a" "b" "cd" "e".
I tried with line.split('";"'), but it did not help.
import csv
reader = csv.reader(open("yourfile.csv", "rb"), delimiter=';')
for row in reader:
print row
Try this out.
import csv
reader = csv.reader(open("yourfile.csv", "rb"), delimiter=';', quoting=csv.QUOTE_NONE )
for row in reader:
print row
This ^^^ if you want quotes preserved
Edit: If you want ';' removed from the field content ('c;d' = 'cd' case) - you may do the post processing on rows returned, something like this:
import csv
reader = csv.reader(open("yourfile.csv", "rb"), delimiter=';', quoting=csv.QUOTE_NONE )
for row in reader:
print [item.replace(';', '') for item in row]
In other contexts, the shlex.split() function could be used

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