This question already has answers here:
How can you dynamically create variables? [duplicate]
(8 answers)
Closed 7 years ago.
I would like to set the variable name in a for loop, like:
for i in range(5):
namei = i # this is a variable name
It will give me:
name0 = 0
name1 = 1
name2 = 2
name3 = 3
name4 = 4
does anyone know how to do that?
Thank you!
Instead of having 5 separate variables, you should use an array.
Eg.
name = [] #Create an empty array
for i in range(5):
name.append(i) #Add each value to your array
This will leave you with
name[0] = 0
name[1] = 1
...
etc.
Related
This question already has answers here:
In Dart, what's the difference between List.from and .of, and between Map.from and .of?
(3 answers)
Closed 1 year ago.
sir, I am working on flutter. I want to know the difference between both codes.
points = List.add(DrawingArea(
point: details.localPosition,
areaPaint: Paint()
..strokeCap = StrokeCap.round
..blendMode = BlendMode.clear
..color = Colors.transparent
..isAntiAlias = true
..strokeWidth = strokeWidth))
points = List.from(points)..add(DrawingArea(
point: details.localPosition,
areaPaint: Paint()
..strokeCap = StrokeCap.round
..blendMode = BlendMode.clear
..color = Colors.transparent
..isAntiAlias = true
..strokeWidth = strokeWidth))
List.add adds a single entry, List.from creates a list from another list (multiples entries)
I have a csv with one of the columns that contains periods:
timespan (string): PnYnMnD, where P is a literal value that starts the expression, nY is the number of years followed by a literal Y, nM is the number of months followed by a literal M, nD is the number of days followed by a literal D, where any of these numbers and corresponding designators may be absent if they are equal to 0, and a minus sign may appear before the P to specify a negative duration.
I want to return a data frame that contains all the data in the csv with parsed timespan column.
So far I have a code that parses periods:
import re
timespan_regex = re.compile(r'P(?:(\d+)Y)?(?:(\d+)M)?(?:(\d+)D)?')
def parse_timespan(timespan):
# check if the input is a valid timespan
if not timespan or 'P' not in timespan:
return None
# check if timespan is negative and skip initial 'P' literal
curr_idx = 0
is_negative = timespan.startswith('-')
if is_negative:
curr_idx = 1
# extract years, months and days with the regex
match = timespan_regex.match(timespan[curr_idx:])
years = int(match.group(1) or 0)
months = int(match.group(2) or 0)
days = int(match.group(3) or 0)
timespan_days = years * 365 + months * 30 + days
return timespan_days if not is_negative else -timespan_days
print(parse_timespan(''))
print(parse_timespan('P2Y11M20D'))
print(parse_timespan('-P2Y11M20D'))
print(parse_timespan('P2Y'))
print(parse_timespan('P0Y'))
print(parse_timespan('P2Y4M'))
print(parse_timespan('P16D'))
Output:
None
1080
-1080
730
0
850
16
How do I apply this code to the whole csv column while running the function processing csv?
def do_process_citation_data(f_path):
global my_ocan
my_ocan = pd.read_csv(f_path, names=['oci', 'citing', 'cited', 'creation', 'timespan', 'journal_sc', 'author_sc'],
parse_dates=['creation', 'timespan'])
my_ocan = my_ocan.iloc[1:] # to remove the first row
my_ocan['creation'] = pd.to_datetime(my_ocan['creation'], format="%Y-%m-%d", yearfirst=True)
my_ocan['timespan'] = parse_timespan(my_ocan['timespan']) #I tried like this, but sure it is not working :)
return my_ocan
Thank you and have a lovely day :)
Like with Python's builtin map, Pandas also has that method. You can check its documentation here. Since you already have your function ready which takes a single parameter and returns a value, you just need this:
my_ocan['timespan'] = my_ocan['timespan'].map(parse_timespan) #This will take each value in the column "timespan", pass it to your function 'parse_timespan', and update the specific row with the returned value
And here is a generic demo:
import pandas as pd
def demo_func(x):
#Takes an int or string, prefixes with 'A' and returns a string.
return "A" + str(x)
df = pd.DataFrame({"Column_1": [1, 2, 3, 4], "Column_2": [10, 9, 8, 7]})
print(df)
df['Column_1'] = df['Column_1'].map(demo_func)
print("After mapping:\n{}".format(df))
Output:
Column_1 Column_2
0 1 10
1 2 9
2 3 8
3 4 7
After mapping:
Column_1 Column_2
0 A1 10
1 A2 9
2 A3 8
3 A4 7
This question already has answers here:
Pandas read_csv dtype leading zeros
(5 answers)
Import pandas dataframe column as string not int
(3 answers)
Closed 5 years ago.
There is a dataframe (df1) like as following after I read the data from txt file:
name l1 l2
a 00000 00000
b 00010 00002
c 00000 01218
When I use the python code as following:
dataframe.to_csv('test.csv', index= False)
Then I use the following code to read:
df = pd.read_csv('test.csv')
I found the dataframe is being df2 as following
name l1 l2
a 0 0
b 10 2
c 0 1218
But I want to keep the leading zero in the dataframe like df1.
Thanks!
The leading zeros are being removed because Pandas is implicitly converting the values to integer types. You want to read the data as string types, which you can do by specifying dtype=str:
pd.read_csv('test.csv', dtype=str)
Update as it helps others:
To have most or selective columns as str, one can do this:
# lst of column names which needs to be string
lst_str_cols = ['prefix', 'serial']
# use dictionary comprehension to make dict of dtypes
dict_dtypes = {x : 'str' for x in lst_str_cols}
# use dict on dtypes
pd.read_csv('sample.csv', dtype=dict_dtypes)
I have a dataset with string variables and I am trying to generate a new binary variable based on the first two characters. All strings are 5 characters long, but I'm only concerned with the first two in order to sort.
For example, I could have 22001 and 22005. Since both are of the form 22XXX, I want to assign value 1 for both in the variable type_A. And if I have 25001 and 25005, since both are not of the form 22XXX, I want to assign value 0 for both in the variable type_A.
This should do the job:
clear
set obs 4
generate str5 var1 = "22001" in 1
replace var1 = "22005" in 2
replace var1 = "25001" in 3
replace var1 = "25005" in 4
gen type_A = substr(var1, 1, 2) == "22"
Please note that as you explain your problem it looks like you you are storing 22005 as text - which may not necessarily be the best idea..
I already loaded 20 csv files with function:
tbl = list.files(pattern="*.csv")
list_of_data = lapply(tbl, read.csv)
I combined all of those filves into one:
all_data = do.call(rbind.fill, list_of_data)
In the new table is a column called "Accession". After combining many of the names (Accession) are repeated. And I would like to remove all of the duplicates.
Another problem is that some of those "names" are ALMOST the same. The difference is that there is name and after become the dot and the number.
Let me show you how it looks:
AT3G26450.1 <--
AT5G44520.2
AT4G24770.1
AT2G37220.2
AT3G02520.1
AT5G05270.1
AT1G32060.1
AT3G52380.1
AT2G43910.2
AT2G19760.1
AT3G26450.2 <--
<-- = Same sample, different names. Should be treated as one. So just ignore dot and a number after.
Tried this one:
all_data$CleanedAccession = str_extract(all_data$Accession, "^[[:alnum:]]+")
all_data = subset(all_data, !duplicated(CleanedAccession))
Error in `$<-.data.frame`(`*tmp*`, "CleanedAccession", value = character(0)) :
You can use this command to both subset and rename the values:
subset(transform(alldata, Ascension = sub("\\..*", "", Ascension)),
!duplicated(Ascension))
Ascension
1 AT3G26450
2 AT5G44520
3 AT4G24770
4 AT2G37220
5 AT3G02520
6 AT5G05270
7 AT1G32060
8 AT3G52380
9 AT2G43910
10 AT2G19760
What about
df <- data.frame( Accession = c("AT3G26450.1",
"AT5G44520.2",
"AT4G24770.1",
"AT2G37220.2",
"AT3G02520.1",
"AT5G05270.1",
"AT1G32060.1",
"AT3G52380.1",
"AT2G43910.2",
"AT2G19760.1",
"AT3G26450.2"))
df[!duplicated(unlist(lapply(strsplit(as.character(df$Accession),
".", fixed = T), "[", 1))), ]