[英]Python: csv - find value in column
Suppose I have a csv file looks like: 假设我有一个csv文件,如下所示:
(a,3,,,,)
(b,,,,,)
(c,2,,,,)
(d,,,,,)
(e,,,,,)
(f,1,,,,)
(g,,,,,)
I am trying to find out which alphabet (ie a,b,c,d,e,f,g) has a value in column (let's say column 1 here). 我试图找出哪一个字母(即a,b,c,d,e,f,g)在列中有一个值(这里说第1列)。
The code I've written is here: 我编写的代码在这里:
set3 = set([n[column] for n in new_marks])
if set3 != '':
print
print '{0} is not empty'.format(list(set3))
This code only prints value in column 1 not alphabets... 此代码仅打印第1列中的值,而不打印字母...
Can anyone help me with figuring out this problem? 谁能帮我解决这个问题?
Thank you 谢谢
I would do something similar to: 我会做类似的事情:
import csv
def strip_bits(obj):
for el in obj:
yield el[1:-2]
with open('/home/jon/testfile.csv') as fin:
tempin = strip_bits(fin)
csvin = csv.reader(tempin)
for row in csvin:
if any(row[1:]):
print row[0] # or do whatever
without csv
solution: 没有csv
解决方案:
with open('data.txt') as f:
for x in f:
x=x.strip().strip('()')
spl=x.split(',')
if spl[0] in ('a','b','c','d','e','f','g') and spl[1]!='': #replace spl[1] with
# any other column if you want to find value in some other column
print ("{0} is not empty".format(spl[0]))
output: 输出:
a is not empty
c is not empty
f is not empty
d = dict((n[0], n[column]) for n in new_marks)
if d:
print
print '{0} is not empty'.format(d.keys())
修改第一行:
set3 = set([n[0] for n in new_marks if n[column] is not None])
this code does pretty much what i understand from your question. 这段代码几乎可以理解您的问题。
i am not sure what you define as a "value" so here it is simply any row that has more than 1 column. 我不确定您将其定义为“值”的原因,因此这里仅仅是具有多于一列的任何行。 but this code can be easily modified. 但是此代码可以轻松修改。
#holds the rows from the file
rows = []
#opens and reads each row of the file (CSVFILE = full file location)
with open(CSVFILE, "r") as CSV_read:
next(CSV_read) # removes headers/colum names (if you dont have any - remove this line)
for line in CSV_read:
#cleans and splits the line into colums
vals = line.strip("\r\n").strip("\n").split(",")
#adds the line to the list of rows
rows.append(vals)
#holds the alphabet (first colum value) of each row that has more values/cells
have_value = []
#goes over each row in the rows
for row in rows:
#if there are any values after the initial 'alphabet'
if len(row) > 1:
# add to the list of things with values.
have_value.append(row[0])
How about this? 这个怎么样?
s = """a,3,,,,
b,,,,,
c,2,,,,
d,,,,,
e,,,,,
f,1,,,,
g,,,,,"""
buf = StringIO(s)
d = {}
for row in csv.reader(buf,delimiter=','):
d[row[0]] = row[1:]
looking_for = 2
for alpha,items in d.iteritems():
try:
if items[looking_for-1] == '1':
print 'Letter %s has value in column %s' % (alpha,looking_for)
except IndexError:
print "List doesn't contain %s columns!" % looking_for
A CSV is nothing but a pure comma-seperated text file, isn't it ? CSV不过是纯逗号分隔的文本文件,不是吗?
with open("/path/to/myfile.csv") as f:
for line in f:
line_chars = line.strip()[1:-1].split(",")
k, v= line_chars[0], "".join(line_chars[1:])
if v: print k, v
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