我编写了一个 for 循环,它迭代 CSV 以获得如下列表:
[t1, s1]
[t2, s2]
[t3, s3]
等等 4000 次。 现在我需要将它们写入一个新的 CSV 文件,其中它们将填充 2 个字段并用逗号分隔。 当我输入此内容时,我只获取最后一个循环中的最后一个列表,并且单元格中包含一个字符。
def sentiment_analysis():
fo = open("positive_words.txt", "r")
positive_words = fo.readlines()
fo.close()
positive_words = map(lambda positive_words: positive_words.strip(), positive_words)
fo = open("negative_words.txt", "r")
negative_words = fo.readlines()
fo.close()
negative_words = map(lambda negative_words: negative_words.strip(), negative_words)
fo = open("BAC.csv", "r")
data = fo.readlines()
fo.close()
data = map(lambda data: data.strip(), data)
x1 = 0 #number of bullish
x2 = 0 #number of bearish
x3 = 0 #number of unknown
for info in data:
data_specs = info.split(',')
time_n_date = data_specs[0]
sentiment = data_specs[2]
'''Possibly precede with a nested for loop for data_specs???'''
if sentiment == 'Bullish':
'''fo.write(time + ',' + 'Bullish' + '\n')'''
elif sentiment == 'Bearish':
''' fo.write(time + ',' + 'Bearish' + '\n')'''
else:
x3 += 1
positive = 0
negative = 0
content_words = data_specs[1].split()
for a in positive_words:
for b in content_words:
if (a == b):
positive = positive + 1
for c in negative_words:
for d in content_words:
if (c == d):
negative = negative + 1
if positive > negative:
'''fo.write(time + ',' + 'Bullish' + '\n')'''
sentiment = 'Bullish'
elif positive < negative:
sentiment = 'Bearish'
else:
sentiment = 'Neutral'
bac2data = [time_n_date, sentiment]
print bac2data
fo = open("C:\Users\Siddhartha\Documents\INFS 772\Project\Answer\BAC2_answer.csv", "w")
for x in bac2data:
w = csv.writer(fo, delimiter = ',')
w.writerows(x)
fo.close()
我的 for 循环没有完成这一切。
最佳答案
在您的代码中bac2data = [time_n_date,情感]
创建一个包含2个字符串项的列表。使用 csv.writer()
将其写入 CSV 文件的正确方法是使用 writerow(bac2data)
。
代码的最后部分包含许多错误。首先,您以写入模式 ('w'
) 打开 CSV 文件,获取每一行传入数据。这将每次覆盖文件,丢失除最后一行之外的所有数据。然后,您将迭代 bac2data
列表并在每个项目上调用 writerows()
。这会将字符串中的每个字符写在它自己的行上(与您报告的输出相匹配)。
相反,打开输出文件并在主for info in data:
循环之外创建一个csv.writer
:
fo = open("C:\Users\Siddhartha\Documents\INFS 772\Project\Answer\BAC2_answer.csv", "w")
writer = csv.writer(fo)
for info in data:
....
然后替换主循环底部的这些行:
bac2data = [time_n_date, sentiment]
print bac2data
fo = open("C:\Users\Siddhartha\Documents\INFS 772\Project\Answer\BAC2_answer.csv", "w")
for x in bac2data:
w = csv.writer(fo, delimiter = ',')
w.writerows(x)
fo.close()
这样:
bac2data = [time_n_date, sentiment]
print bac2data
writer.writerow(bac2data)
一旦您可以使用该功能,并且不再需要打印 bac2data
进行调试,您只需使用 1 行即可:
writer.writerow((time_n_date, sentiment)]
<小时/>
更新
函数的完整代码:
def sentiment_analysis():
fo = open("positive_words.txt", "r")
positive_words = fo.readlines()
fo.close()
positive_words = map(lambda positive_words: positive_words.strip(), positive_words)
fo = open("negative_words.txt", "r")
negative_words = fo.readlines()
fo.close()
negative_words = map(lambda negative_words: negative_words.strip(), negative_words)
fo = open("BAC.csv", "r")
data = fo.readlines()
fo.close()
data = map(lambda data: data.strip(), data)
x1 = 0 #number of bullish
x2 = 0 #number of bearish
x3 = 0 #number of unknown
fo = open("C:\Users\Siddhartha\Documents\INFS 772\Project\Answer\BAC2_answer.csv", "w")
writer = csv.writer(fo)
for info in data:
data_specs = info.split(',')
time_n_date = data_specs[0]
sentiment = data_specs[2]
'''Possibly precede with a nested for loop for data_specs???'''
if sentiment == 'Bullish':
'''fo.write(time + ',' + 'Bullish' + '\n')'''
elif sentiment == 'Bearish':
''' fo.write(time + ',' + 'Bearish' + '\n')'''
else:
x3 += 1
positive = 0
negative = 0
content_words = data_specs[1].split()
for a in positive_words:
for b in content_words:
if (a == b):
positive = positive + 1
for c in negative_words:
for d in content_words:
if (c == d):
negative = negative + 1
if positive > negative:
'''fo.write(time + ',' + 'Bullish' + '\n')'''
sentiment = 'Bullish'
elif positive < negative:
sentiment = 'Bearish'
else:
sentiment = 'Neutral'
bac2data = [time_n_date, sentiment]
print bac2data
writer.writerow(bac2data)
fo.close()
关于python - 将 for 循环中的列表写入 csv,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35834654/